Download Math 55b Lecture Notes Contents

Math 55b Lecture Notes
Evan Chen
Spring 2015
This is Harvard College’s famous Math 55b, instructed by Dennis Gaitsgory.
The formal name for this class is “Honors Real and Complex Analysis” but it
generally goes by simply “Math 55b”. It is an accelerated one-semester class
covering the basics of analysis, primarily real but also some complex analysis.
The permanent URL is http://web.evanchen.cc/~evanchen/coursework.
html, along with all my other course notes.
Special thanks to W Mackey for providing notes on the several day that I
slept through class.
Contents
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5
January 29, 2015
1.1 Definition and Examples . . . . . . . . . . .
1.2 Continuous Maps . . . . . . . . . . . . . . .
1.3 Forgetting the Metric – Topological Spaces
1.4 Intermediate Value Theorem . . . . . . . .
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February 3, 2015
2.1 Sequential Continuity . .
2.2 Cauchy Completeness . .
2.3 R is Complete . . . . . .
2.4 1-countable and Hausdorff
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February 5, 2015
3.1 Completion . . . . . . .
3.2 Why no compactness . .
3.3 Sequential Compactness
3.4 Compactness . . . . . .
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February 12, 2015
4.1 Warm-Up . . . . . . .
4.2 “Applied” Math . . .
4.3 Series . . . . . . . . .
4.4 Convergent Things . .
4.5 Convergence Tests . .
4.6 The Series n−a . . . .
4.7 Exponential Function
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February 17, 2015
22
1
5.1 L norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
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Evan Chen (Spring 2015)
5.2
5.3
5.4
5.5
5.6
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Contents
Defining the integral .
Uniform Continuity .
Defining the Integral .
Properties of Integrals
Riemann Sums . . . .
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March 3, 2015
9.1 Differentiable forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.3 Higher-Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . .
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February 19, 2015
6.1 Bonus Problem Sets
6.2 PSet 2, Problem 7b
6.3 PSet 3, Problem 8 .
6.4 PSet 3, Problem 10
6.5 PSet 3, Problem 4 .
6.6 Filters . . . . . . . .
6.7 PSet 4 . . . . . . . .
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February 24, 2014
7.1 Limits . . . . . . . . .
7.2 Differentiation . . . .
7.3 Chain Rule . . . . . .
7.4 More “applied math”
7.5 Higher Derivatives . .
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February 26, 2015
8.1 Fundamental Theorem of Calculus
8.2 Pointwise converge sucks . . . . .
8.3 Uniform Convergence of Functions
8.4 Applied Math – Differentiating ex
8.5 Differentiable Paths in Rn . . . . .
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10 March 5, 2015
10.1 Inverse function theorem . . . . . . . .
10.2 Proof of the Inverse Function Theorem .
10.3 Step 0 . . . . . . . . . . . . . . . . . . .
10.4 First Step . . . . . . . . . . . . . . . . .
10.5 Third Step . . . . . . . . . . . . . . . .
10.6 Second Step . . . . . . . . . . . . . . . .
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11 March 10, 2015
11.1 Completing the Proof of Inverse Function Theorem
11.2 Implicit Function Theorem . . . . . . . . . . . . .
11.3 Differential Equations . . . . . . . . . . . . . . . .
11.4 Lawns . . . . . . . . . . . . . . . . . . . . . . . . .
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12 March 12, 2015
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13 March 24, 2015
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13.1 Midterm Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2
Evan Chen (Spring 2015)
Contents
13.2 PSet Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
13.3 Review of Exterior Products . . . . . . . . . . . . . . . . . . . . . . . . . 59
13.4 Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
14 March 26, 2015
14.1 More on Differential Forms
14.2 Topological Manifolds . . .
14.3 Smoothness of Manifolds .
14.4 Cotangent space . . . . . .
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15 March 31, 2015
15.1 Boxes . . . . . . . . . . . . . .
15.2 Supports . . . . . . . . . . . .
15.3 Partitions of Unity . . . . . . .
15.4 Existence of Partitions of Unity
15.5 Case 1: A is Compact . . . . .
15.6 Case 2: A is a certain union . .
15.7 Case 3: A open . . . . . . . . .
15.8 Case 4: General A . . . . . . .
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16 April 2, 2015
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16.1 Integration on Bounded Open Sets . . . . . . . . . . . . . . . . . . . . . . 71
16.2 This Integral is Continuous . . . . . . . . . . . . . . . . . . . . . . . . . . 72
16.3 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
17 April 7, 2015
17.1 Integration Over Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . .
17.2 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17.3 Integration Over Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . .
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18 April 9, 2015
18.1 Digression: Orientations on Complex
18.2 Setup . . . . . . . . . . . . . . . . .
18.3 Integration on Manifolds . . . . . . .
18.4 Stoke’s Theorem . . . . . . . . . . .
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Vector Spaces
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19 April 14, 2015
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20 April 16, 2015
20.1 Complex Differentiation . . . . . .
20.2 Examples . . . . . . . . . . . . . .
20.3 Inverse Function Theorem . . . . .
20.4 Complex Differentiable Forms . . .
20.5 Cauchy Integral Formula . . . . .
20.6 Computation when f = 1 . . . . .
20.7 General Proof of Cauchy’s Formula
21 April 21, 2015
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22 April 28, 2015
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22.1 Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 96
22.2 Proofs of First Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
3
Evan Chen (Spring 2015)
Contents
22.3 Proof of Hurwitz’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 98
22.4 A Theorem We Tried to Use But Couldn’t . . . . . . . . . . . . . . . . . 99
4
Evan Chen (Spring 2015)
1 January 29, 2015
§1 January 29, 2015
Grading system for 55B with ' 55A.
We’ll be more or less following Baby Rudin.
§1.1 Definition and Examples
This subsection is copied from my Napkin project.
Definition 1.1. A metric space is a pair (M, d) consisting of a set of points M and a
metric d : M × M → R≥0 . The distance function must obey the following axioms.
• For any x, y ∈ M , we have d(x, y) = d(y, x); i.e. d is symmetric.
• The function d must be positive definite which means that d(x, y) ≥ 0 with
equality if and only if x = y.
• The function d should satisfy the triangle inequality: for all x, y, z ∈ M ,
d(x, z) + d(z, y) ≥ d(x, y).
Abuse of Notation 1.2. Just like with groups, we will abbreviate (M, d) as just M .
Example 1.3 (Metric Spaces of R)
(a) The real line R is a metric space under the metric d(x, y) = |x − y|.
(b) The interval [0, 1] is also a metric space with the same distance function.
(c) In fact, any subset S of R can be made into a metric space in this way.
Example 1.4 (Metric Spaces of R2 )
(a) We can make R2 into a metric space by imposing the Euclidean distance
function
p
d ((x1 , y1 ), (x2 , y2 )) = (x1 − x2 )2 + (y1 − y2 )2 .
(b) Just like with the first example, any subset of R2 also becomes a metric space
after we inherit it. The unit disk, unit circle, and the unit square [0, 1]2 are
special cases.
Example 1.5 (Taxicab on R2 )
It is also possible to place the following taxicab distance on R2 :
d ((x1 , y1 ), (x2 , y2 )) = |x1 − x2 | + |y1 − y2 | .
For now, we will use the more natural Euclidean metric. (One can also use max
instead of a sum.)
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Example 1.6 (Metric Spaces of Rn )
We can generalize the above examples easily. Let n be a positive integer. We define
the following metric spaces.
(a) We let Rn be the metric space whose points are points in n-dimensional
Euclidean space, and whose metric is the Euclidean metric
p
d ((a1 , . . . , an ) , (b1 , . . . , bn )) = (a1 − b1 )2 + · · · + (an − bn )2 .
This is the n-dimensional Euclidean space.
(b) The open unit ball B n is the subset of Rn consisting of those points (x1 , . . . , xn )
such that x21 + · · · + x2n < 1.
(c) The unit sphere S n−1 is the subset of Rn consisting of those points (x1 , . . . , xn )
such that x21 + · · · + x2n = 1, with the inherited metric. (The superscript n − 1
indicates that S n−1 is an n − 1 dimensional space, even though it lives in
n-dimensional space.) For example, S 1 ⊆ R2 is the unit circle, whose distance
between two points is the length of the chord joining them. You can also think
of it as the “boundary” of the unit ball B n .
Example 1.7 (Discrete Space)
Let S be any set of points (either finite or infinite). We can make S into a discrete
space by declaring the following distance function.
(
1 if x 6= y
d(x, y) =
0 if x = y.
If |S| = 4 you might think of this space as the vertices of an regular tetrahedron,
living in R3 . But for larger S it’s not so easy to visualize. . .
Example 1.8 (Graphs are Metric Spaces)
Any connected simple graph G can be made into a metric space by defining the
distance between two vertices to be the graph-theoretic distance between them. (The
discrete metric is the special case when G is the complete graph on S.)
Example 1.9 (Function Space)
We can let MR be the space of integrable functions [0, 1] → R and define the metric
1
by d(f, g) = 0 |f − g| dx.
§1.2 Continuous Maps
This is again largely excerpted from my Napkin project.
Abuse of Notation 1.10. For a function f and its argument x, we will begin abbreviating f (x) to just f x when there is no risk of confusion.
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In calculus you were also told (or have at least heard) of what it means for a function
to be continuous. Probably something like
A function f : R → R is continuous at a point p ∈ R if for every ε > 0 there
exists a δ > 0 such that |x − p| < δ =⇒ |f x − f p| < ε.
Question 1.11. Can you guess what the corresponding definition for metric spaces is?
All we have do is replace the absolute values with the more general distance functions:
this gives us a definition of continuity for any function M → N .
Definition 1.12. Let M = (M, dM ) and N = (N, dN ) be metric spaces. A function
f : M → N is continuous at a point p ∈ M if for every ε > 0 there exists a δ > 0 such
that
dM (x, p) < δ =⇒ dN (f x, f p) < ε.
Moreover, the entire function f is continuous if it is continuous at every point p ∈ M .
Notice that, just like in our definition of an isomorphism of a group, we use both the
metric of M for one condition and the metric of N on the other condition.
Example 1.13
Let M be any metric space and D a discrete space. When is a map f : D → M
continuous? Any map D → M is continuous.
Proof. Take an open ball of radius 12 .
Example 1.14
The map R → R by x 7→ x2 is continuous. So is x 7→ x3 .
Proof. Homework.
Example 1.15
Let X = Rn with one product metric and let Y = Rn with another product metric.
Then id : X → Y is continuous. Thus we will generally not be pedantic about the
choice of metric.
§1.3 Forgetting the Metric – Topological Spaces
Again excerpted from the Napkin project.
Definition 1.16. A topological space is a pair (X, T ), where X is a set of points, and
T is the topology, which consists of several subsets of X, called the open sets of X.
The topology must obey the following four axioms.
• ∅ and X are both in T .
• Finite intersections of open sets are also in T .
• Arbitrary unions (possibly infinite) of open sets are also in T .
Abuse of Notation 1.17. We refer to the space (X, T ) by just X. (Do you see a
pattern here?)
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Example 1.18
We can declare all sets are open: a discrete space is a topological space in which
every set is open.
Example 1.19
Declare U open if and only if ∀x ∈ U , the ball of radius r centered at x is contained
in U .
Proof. Check all the axioms. Blah.
Proposition 1.20
Let M be a metric space. Then for all x ∈ M and r > 0, the ball
B(x, r) = {y | dM (x, y) < r}
is open.
Proof. Pick y in the ball. Let t = d(x, y). Then t < r, so pick ε with t + ε < r. You can
check using the triangle inequality that B(y, ε) ⊆ B(x, r).
Example 1.21
[0, 1] is not open since no ball at 0 is contained inside it.
Definition 1.22. A subset Y is closed iff X − Y is open.
Definition 1.23. A function f : X → Y of topological spaces is continuous at p ∈ X
if the pre-image of any neighborhood of f p is also a neighborhood of p.
With some effort, we can show this is the same definition of continuity as with metric
spaces.
§1.4 Intermediate Value Theorem
Theorem 1.24 (IVT)
Let f : [0, 1] → R be continuous such that f (0) < 0 and f (1) > 0. Then ∃a ∈ [0, 1]
such that f (a) = 0.
This theorem is not cheap, and requires the following theorem.
Theorem 1.25
Let A ⊆ R be bounded above. Then there exists a least upper bound y ∈ R.
Proof of IVT. Long and boring. Just draw a picture. The main point is to take y to be
the least upper bound of the a for which f (a) ≤ 0.
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“This is just a game of quantifiers. To do this, all you have to do is be sober,
which is not a problem since you are all under 21.
. . . I cannot do this past 6PM, not because I’m not sober, but because I am
old.”
– Dennis Gaitsgory
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§2 February 3, 2015
Didn’t attend class.
§2.1 Sequential Continuity
Continuous functions send convergent sequences to convergent sequences, limits sent to
limits.
The converse is true in metric spaces.
§2.2 Cauchy Completeness
So far we can only talk about sequences converging if they have a limit.√ But consider the
sequence x1 = 1, x2 = 1.4, x3 = 1.41, x4 = 1.414, . . . . It converges to 2 in R, of course.
But it fails to converge in Q. And so somehow, if we didn’t know about the existence of
R, we would have no idea that the sequence (xn ) is “approaching” something.
That seems to be a shame. Let’s set up a new definition to describe these sequences
whose terms eventually get close to each other, but don’t necessarily converge to a point.
Definition 2.1. Let x1 , x2 , . . . be a sequence which lives in a metric space M = (M, dM ).
We say it is Cauchy if for any ε > 0, we have
dM (xm , xn ) < ε
for all sufficiently large m and n.
Note that, unlike the rest of this chapter, this is a notion which applies only to metric
spaces. In a general topological space there is not a good enough notion of “distance” to
make this definition work.
Question 2.2. Show that a sequence which converges is automatically Cauchy. (Draw
a picture.)
Now we can define the following.
Definition 2.3. A metric space M is complete if every Cauchy sequence converges.
Most metric spaces aren’t complete, like Q. But it turns out that every metric space can
be completed by “filling in the gaps” somehow, resulting in a space called the completion
of the metric space. The construction is left as an (in my opinion) fun problem.
It’s a theorem that R is complete. To prove this I’d have to define R rigorously, which
I won’t do here (yet). In fact, there are some competing definitions of R. It is sometimes
defined as the completion of the space Q. Other times it is defined using something called
Dedekind cuts. For now, let’s just accept that R behaves as we expect and is complete.
Example 2.4 (Examples of Complete Sets)
(a) R is complete.
(b) The discrete space is complete, as the only Cauchy sequences are eventually
constant.
(c) The closed interval [0, 1] is complete.
(d) Rn is in fact complete as well. (You are welcome to prove this by induction on
n.)
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Example 2.5 (Non-Examples of Complete Sets)
(a) The rationals Q are not complete.
(b) The open interval (0, 1) is not complete, as the sequence xn =
does not converge.
1
n
is Cauchy but
§2.3 R is Complete
Theorem 2.6 (Bolzano-Weirestraß)
Any sequence in [0, 1] has a convergent subsequence.
§2.4 1-countable and Hausdorff
Most of the “nice” properties of metric spaces carry over to 1-countable Hausdorff general
topological subspaces.
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§3 February 5, 2015
Last time we defined Cauchy sequences. Cool.
Today we will complete a metric space!
Theorem 3.1
Let X be a metric space.
(a) There exists an X and an isometric embedding ι : X → X so that X is
complete and im ι is dense in X.
(b) Any isometric embedding f : X → Y with Y complete factors uniquely
through X via an isometric embedding. (In other words, X is universal with
this property.)
(c) In (b), if f “(X) is dense than f is an isometry.
Here a dense set X 0 ⊆ X means that every neighborhood of X contains a point of X 0 .
Also, f “(X) just means {f (x) | x ∈ X}.
Corollary 3.2
The completion of Q is R.
Proof. We have a dense isometry Q ,→ R.
Okay, it’s not actually that straightforward, you need some properties of R. . .
We will assume in what follows that R is complete.
§3.1 Completion
Consider the set of all Cauchy sequncees of X. We define a metric on them by
ρ ({xn } , {yn }) = lim (ρ(xn , yn )) .
n
Actually, we first need to show that this limit exists.
Lemma 3.3
If {xn }, {yn } are Cauchy, then the sequence ρ(xn , yn ) ∈ R as n = 1, 2, . . . is Cauchy.
Proof. By “Quadrilateral Inequality”,
|ρ(xn , yn ) − ρ(xN , yN )| ≤ ρ(xn , xN ) + ρ(yn , yN ).
Using the ε/2 trick completes the proof.
Assuming that R is complete (COUGH COUGH), this now implies that limn (ρ(xn , yn ))
exists. You can check it also satisfies the Triangle Inequality.
Unfortunately, it’s easy to find examples with ρ({xn }, {yn }) = 0, with (xn ) 6= (yn ). So
we “mod out” by this.
Hence, we define C(X) as the set of all Cauchy sequences modded out by the relation
xn ∼ yn if ρ({xn }, {yn }) = 0. The rest is left as homework.
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§3.2 Why no compactness
Lemma 3.4
A function f : [0, 1] → R is bounded above.
Proof. Otherwise we get a sequence of points x1 , x2 , . . . such that f (xm ) > m for all
m. Then we can find a convergent subsequence using Bolzano-Weirestraß. This breaks
sequential continuity.
Theorem 3.5
Any function f : [0, 1] → R has a global maximum.
Proof. By the lemma, f “[0, 1] is bounded above and we can take the least upper bound y.
We claim this is actually in the limit. If not, then we can construct a sequence x1 , x2 , . . .
1
such that f (xm ) ≥ y − m
. Take a convergent subsequence by Bolzano-Weirestraß. Then
(xm ) converges to some x ∈ [0, 1], but then we must have f (x) = y.
This sucks. Compactness is better. Rawr.
Second Proof, from Jane. As before, there is a least upper bound y. Assume for contradiction that the bound is never obtained. Then the function [0, 1] → R by
x 7→
1
y − f (x)
is an unbounded continuous function on [0, 1], impossible.
§3.3 Sequential Compactness
Definition 3.6. A topological space is sequentially compact if every sequence has a
convergent subsequence.
Proposition 3.7
Let f : X → Y . If X is sequentially compact then f (X) is sequentially compact.
Proof. Trivial. If (f xn ) is a sequence, take a converge subsequence of (xn ) ∈ X.
Theorem 3.8 (Heine-Borel)
A ⊆ R is sequentially compact if and only if it closed and bounded.
We can now kill the original maximum theorem. Given f : [0, 1] → R, its image in R is
bounded and closed and we can easily use this to show that f has a maximum.
To prove Heine-Borel, we first prove one direction in greater generality.
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Lemma 3.9
Let Y 0 ⊆ Y , where Y 0 is sequentially compact and Y is 1-countable and Hausdorff.
Then Y 0 is closed in Y .
Proof. Because Y is 1-countable, we can use the sequence definition of closed in the tricky
direction: it suffices to prove that if y1 , y2 , · · · → y with yi ∈ Y 0 then y ∈ Y 0 . But yi has
a convergent subsequence to some y 0 . It had also better converge to y. By Hausdorff,
y = y0.
Lemma 3.10
Suppose Y 0 ⊆ Y for some topological spaces, with Y a Hausdorff space. If Y 0 is
closed and Y is sequentially compact then Y 0 is sequentially compact.
Proof. Triviality: given yn in Y 0 use sequential compactness of Y to force some subsequence to converge to y ∈ Y . Since Y 0 is closed, y ∈ Y 0 .
Proof of Heine-Borel. First, suppose A is bounded and closed; by bounded-ness it lives
in some A ⊆ [a, b]. By Bolzano-Weirestraß, the set [a, b] is sequentially compact. By our
lemma, A ⊆ [a, b] is sequentially compact.
For the converse, suppose A is sequentially compact. We did a lemma that show A is
closed. Hence A is bounded.
§3.4 Compactness
Apparently Gaitsgory does not like sequential compactness QQ.
Definition 3.11. A topological space is compact if every open cover has a finite subcover.
Proposition 3.12
If f : X → Y is continuous and Y is compact then f (X) is compact.
Proof. Tautological using the definition of continuity.
We can mirror this for sequential compactness now.
Lemma 3.13
Let Y be a Hausdorff space and Y 0 a compact subspace. Then Y 0 is closed.
Proof. We will show that for any y ∈ Y \ Y 0 , there is a neighborhood U ⊆ Y \ Y 0
containing y. For every y 0 ∈ Y 0 , we can use the Hausdorff condition to find neighborhoods
Uy0 3 y 0 and Vy0 3 y. By definition, Y 0 is covered by Uy0 , and we can find a finite subcover
Y0 ⊆
Now take
n
\
i=1
n
[
Uyi
i=1
Vyi 3 y.
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This is a neighborhood of y ∈ Y disjoint from Y 0 .
Proposition 3.14
If X is first-countable and X is compact, then it is sequentially compact.
Proof. Suppose not. Let {xn } be a sequence with no convergent subsequence. Using
1-countable, we find that for every x ∈ X, there exists a Ux contains
S only finitely many
elements of the sequence. But then we can take a finite subcover Ux .
Proposition 3.15
If X is second-countable, then sequentially compact implies compact.
Here second countable means there is a countable base.
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§4 February 12, 2015
“Today we will be doing applied math” – Gaitsgory
“This can’t be happening”
“Applied by my standards”.
§4.1 Warm-Up
Lemma 4.1
Let ai , bi be sequences converging to a and b. Then ai + bi → a + b, ai bi → ab.
Proof. The maps +, × : R2 → R are continuous.
§4.2 “Applied” Math
Also known as: cooking up stupid bounds.
Also known as: suppose an ≤ xn ≤ bn for all n, where an and bn are the locations of
two cops and xn is the location of a drunkard.
Now a surprisingly nontrivial statement.
Example 4.2
Let |a| < 1. Then the sequence (an )n converges to 0.
Proof. WLOG a > 0 (the other case is easy). We bound the sequence an between two
guys. Put the estimate
n
1
1
≥n
−1 .
a
a
From this we deduce
0 ≤ an ≤
1
a
1
1
· .
−1 n
Example 4.3
√
For any real number a > 0, the sequence ( n a)n converges to 1.
The existence of
√
n
a follows from the Intermediate Value Theorem.
Proof. It suffices to consider a > 1, since otherwise we can consider 1/a. Let xn be such
that xnn = a for each n. Then
a n
1≤a< 1+
n
so 1 ≤ x ≤ 1 + na .
Example 4.4
√
The sequence ( n n)n converges to 1.
Proof. 1 ≤ n ≤ (1 + n1 )n by Bernoulli’s inequality.
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§4.3 Series
We’ll put curly brackets around series for emphasis.
Definition 4.5. A series {an } converges if the sequence bn = a1 + · · · + an converges.
Here is a better notion.
Definition 4.6. The series {an } converges absolutely if bn = |a1 |+· · ·+|an | converges.
Example 4.7
For |a| < 1 the series {an }n converges absolutely.
Proof. 1 + · · · + an =
1−an+1
1−a
→
1
1−a .
Lemma 4.8
If the series {an } converges then the sequence (an ) converges to 0.
Remark 4.9. The converse is false; for example, take the harmonic series.
Proof. The partial sums Sn converge as sequences to some b. Then (Sn+1 − Sn )n is a
convergent sequence to b − b = 0.
Lemma 4.10
The series {an } converges if and only if ∀ε > 0 there is a natural N such that
n
2
X
ai < ε
i=n1
for all n1 , n2 > N .
Proof. The condition just says that the partial sums are a Cauchy sequence. Since R is
complete, that’s equivalent to partial sums converging.
Lemma 4.11
A series converging absolutely also converges.
Proof. Use the triangle inequality in the previous lemma in the form
X
X ε>
|ai | ≥ ai .
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§4.4 Convergent Things
Theorem 4.12
Consider a convergent series {an }.
(a) If {an } converges absolutely, then for any permutation σ : N → N, the series
{aσ(n) }n converges absolutely to the same value.
(b) If {an } converges but not absoluetly, then for any L ∈ R we can permute the
sequence σ : N → N so that {aσ(n) } converges to L.
This is why absolutely convergent series are better than non-convergent series.
Lemma 4.13
A monotonically increasing sequence bn converges if and only if it’s bounded above.
Proof. One direction’s clear. Now assume it’s increasing and bounded by L. By compactness on [b1 , L] we get a convergent subsequence and we’re happy.
§4.5 Convergence Tests
Proposition 4.14
Let {an } have positive terms. Then {an } converges if and only if there exists A such
that
n
X
an ≤ A
i=1
for all n.
Proof. Let Sn be the partial sums and apply the previous lemma.
Corollary 4.15
Let {an } and {bn } have positive terms. If all partial sums of an are at most the
partial sums of bn , then convergence of bn implies convergence of an .
Proposition 4.16 (Cauchy Root Test)
Let {an } be a series.
p
n
|an | < 1. Then {an } converges absolutely.
p
(b) Assume lim sup n |an | > 1. Then {an } diverges.
(a) Assume lim sup
Proof. For part (a), let a = lim sup
many n’s,
p
n
|an |. Let ε such that a + ε < 1. For all but finitely
√
n
an ≤ a + ε.
So for sufficiently large n we have an ≤ (a + ε)n , done.
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For part (b), note that infinitely many terms are actually greater than 1 which is
impossible.
No conclusion when the lim sup equals 1.
Proposition 4.17 (Ratio Test)
Let {an } be a series and assume that
an+1 < 1.
lim sup an Then {an } converges absolutely.
Proof. Let r be the lim sup and let ε > 0 such that r + ε < 1. For sufficiently large n we
have |an+1 | < (r + ε) |an | and bound the sequence above.
Example 4.18
The series {an } given by an =
xn
n!
converges absolutely.
Proof. Direct application of ratio test.
§4.6 The Series n−a
In homework, we will define what xc means for real numbers x and c. In fact I can tell
you:
def
xc = q→c
lim xq .
q∈Q
n−c ,
Now let’s consider the sequence an =
where c > 0 is a real number. The Ratio and
Root tests both fail. Here is the answer.
Theorem 4.19 (Zeta Series)
The sequence {an } given by an = n−c converges for c > 1 and diverges for c ≤ 1. In
particular the harmonic series
1 1 1
+ + + ...
1 2 3
is divergent.
The idea is the “powers of two” estimate, which we prove in full generality here.
Lemma 4.20
Let {an } be a series of positive real numbers which is monotically decreasing, and let
bn = 2n · a2n .
Then {bn } converges if and only if {an } converges.
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The lemma immediately implies the theorem on zeta series.
§4.7 Exponential Function
We define
∞
X
xn
def
exp(x) =
n=0
n!
.
We showed earlier that this always converges.
Lemma 4.21 (Physicist’s Lemma)
Let {an } and {bn } be absolutely convergent series. Define
cn =
n
X
ai bn−i .
i=0
Then {cn } converges to the product of
P
an ·
P
bn .
Proof. We want to show that there for an ε > 0 there exists an N such that when n ≥ N
2n
X
i=0
ci − AB < ε
where A and B are sums of {an } and {bn }.
For sufficiently large N ,
! n
!
n
X
1
X
an
bn − AB ≤ ε.
3
i=0
i=0
So we want to estimate the value of
2n
X
ci −
n
X
i=0
!
an
i=0
n
X
i=0
!
bn .
Expanding, we find that the quantity is actually
X
X
X
X
X
X
a
b
+
a
b
=
b
a
+
b
a
i
j
i
j
j
i
i
j
n<i≤2n
n<j≤2n
1≤j≤n
n<i≤2n
1≤i≤n n<j≤2n
j≤n
i≤n
So it suffices to prove that for large enough n we have
X
X
1
bj
ai < ε.
1≤j≤n n<i≤2n 3
Using absolute convergency,
X
1≤j≤n
|bj |
X
n<i≤2n
|ai |
The left term is at most B, while the right term can be made to be at most
desired.
ε/1000
B ,
as
You can actually weaken the condition to just one series being absolutely convergent.
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Corollary 4.22
exp(x) exp(y) = exp(x + y).
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§5 February 17, 2015
Office hours are now 4PM on Friday.
Today we’ll define an integral. It might seem a bit counter-intuitive, but it is compatible
with reality: you can integrate almost anything, but it’s much harder to differentiate
things.
§5.1 L1 norm
We can consider the set Funccont (X, R) (also called the “max norm” or “L∞ ” or C(X))
of continuous functions from a sequentially compact space X to the real numbers. As
functions of this type attain a maximum value, we can define
ρ(f, g) = max |f (x) − g(x)|
x∈X
and it’s easy to check that this is a norm. There will be plenty of homework problems on
this.
We will mostly be concerned with the space X = [a, b] for this lecture.
§5.2 Defining the integral
We want to define the integral as a continuous map
R
Funccont ([a, b], R) −
→ R.
Of course we already know how to do this for piecewise linear functions, those that are
linear over finitely many intervals of [a, b]. Indeed, we define the integral as the sum of
the linear pieces, where for a linear function f (x) = cx + d we define
Z
b2 − a2
· c.
f = (b − a)d +
2
[a,b]
Remark 5.1. This takes several minutes to actually compute correctly.
So we want to do an extension
R
- R
Funccont ([a, b], R) .................
-
6
R
∪
Funcpiecewise ([a, b], R)
To do so we first need to define uniform continuity.
§5.3 Uniform Continuity
Definition 5.2. Let f : X → Y be a map between metric spaces. We say that f is
uniformly continuous if for all ε > 0 there exists a δ > 0 such that
ρ(x, y) < δ =⇒ ρ(f x, f y) < ε.
This difference is that given an ε > 0 we must specify a δ which works for every input
X. Also, this definition can’t be ported to a general metric space.
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Example 5.3
The functions (0, 1) → R by x 7→ x−1 , x 7→ x2 are not uniformly continuous.
Example 5.4
If f : X → Y satisfies a Lipschitz condition
ρ(f x, f y) ≤ Cρ(x, y)
then f is uniformly continuous.
Remark 5.5. For a differentiable function this is equivalent to saying that the derivatives
are bounded.
Proposition 5.6
Suppose we have a diagram
f
- Y
X ....................
6
f0
ι
X0
so that ι is a dense embedding, f 0 is uniformly continuous and Y is a complete
metric space. Then the map factors uniquely through f as above, and moreover f is
also uniformly continuous.
Proof. We want to define f (x) for x ∈ X. Choose a sequence (xn ) ∈ X 0 converging to X
(possible since the embedding is dense), and consider the sequence f 0 (x1 ), . . . . We claim
that it converges. Indeed, {xn } is Cauchy, so {f 0 (xn )} is Cauchy by directly applying
uniform continuity, and hence continuous by completeness of Y .
Hence for any x ∈ X we define f (x) to be a limit of such a sequence; it’s not hard to
check that this is well-defined.
Uniform continuity is easy but annoying to check. Uniqueness was done on the problem
set.
§5.4 Defining the Integral
We want to apply the proposition to our commutative diagram
R
- R
Funccont ([a, b], R) .................
-
6
R
∪
Funcpiecewise ([a, b], R)
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Evan Chen (Spring 2015)
5 February 17, 2015
To do this we need to check three conditions. First, R is complete, which we already
know. Then, we need to check that the integral is uniformly continuous for piecewise
linear continuous functions. In fact we claim it is Lipschitz : for piecewise linear functions,
we have
Z
Z
f−
g < (b − a) · max |f (x) − g(x)| .
x
[a,b]
[a,b] This is pretty obvious. I won’t bother you.
Now for the actually hard part.
Theorem 5.7
Funcpiecewise ([a, b], R) is dense in Funccont ([a, b], R).
This looks really annoying. Indeed, it would be without the following result.
Theorem 5.8
Let X be a compact metric space and let f : X → Y be continuous where Y is
another metric space. Then f is uniformly continuous. In other words, for compact
sources continuity and uniform continuity are equivalent.
In practice we will take X = [a, b].
Proof. We’re given ε > 0 we want a universal δ. For every x ∈ X let δx be such that
ρX (x0 , x) < δx implies ρY (f x, f x0 ) < 21 ε. Thus we have an open cover
X=
[
x
1
B(x, δx ).
2
1
By compactness we can take a finite subcover. Now let δ = 2015
min δx for those x in the
finite subcover. Consequently, given x1 , x2 ∈ X a distance less than δ from each other,
we see they are both inside some ball centered at some x with radius δx . (Specifically,
take any x which contains x1 in its 12 δx ball.) Then
1
1
ρY (f x1 , f x2 ) ≤ ρ(f x1 , f x) + ρ(f x, f x2 ) < ε + ε = ε.
2
2
Proof of density. Recall that [a, b] is compact. Hence given f : [a, b] → R continuous we
may assume it is uniformly continuous.
Given ε > 0, suffices to approximate our f : [a, b] → R within ε by some f˜ ∈
1
Funcpiecewise ([a, b], R). Choose δ required by uniform continuity of f to get within 2015
ε
in f , and divide [a, b] into a mesh
a = x0 < x1 < · · · < xn−1 = xn = b
with |xk+1 − xk | < δ for each k. Now we do what biologists (and high school hooligans)
do: connect our plot by straight lines. This gives us our f˜, and we can check that it
works: any x ∈ [a, b] is within δ of an xk which is exactly correct.
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Evan Chen (Spring 2015)
5 February 17, 2015
§5.5 Properties of Integrals
Lemma 5.9
R
R
In what follows abbreviate [a,b] to .
R
R
R
(a) f1 + f2 = f1 + f2
R
R
(b) cf = c f .
R
R
R
(c) [a,b] f = [a,c] f + [c,b] f .
R
(d) f ≥ 0 =⇒ f ≥ 0.
(e) Equality holds in the previous point if and only if f = 0.
R
Proof. The point is is a unique extension. The identities hold for piecewise linear
functions; hence they hold for arbitrary continuous functions.
More precisely, we have
R
R
c f − cf
- R
Funcpiecewise ([a, b], R)
...
.
.
.
∩
.
...
.
.
.
.
.
....
......
.
.
.
.
.
.....
......
.
.
.
.
.
? ....
Funccont ([a, b], R)
This resolves (b) and (c). For (a), we instead deal with two maps
R
(f1 +f2 )
Funccont ([a, b], R) × Funccont ([a, b], R) R −→
R
f 1 + f2
R.
For (d) and (e), we can work in
R
Funcpiecewise ([a, b], R≥0 )
- R
...
.....
.
.
.
.....
.....
.
.
.
.
....
.....
.
.
.
.
? ...
Funccont ([a, b], R≥0 )
∩
showing again the embedding is dense.
To prove (e), suppose x ∈ [a, b] such that f (x) = r > 0. Pick an δ > 0 so that
1
[x − δ, x + δ] so that f (x) > 1 − 2015
r > 0 on this interval. From this you have an interval
with area greater than zero.
§5.6 Riemann Sums
Define a mesh p by
a = x0 < x1 < · · · < xn−1 = xn = b
with distinguished points x0i ∈ [xi−1 , xi ] for i = 1, . . . , n. We define
X
Sp (f ) =
(xi − xi−1 ) f (x0i ).
Its width is maxi=1....,n (xi − xi−1 ).
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Evan Chen (Spring 2015)
5 February 17, 2015
Theorem 5.10 (Riemann)
LetRpn be a sequence of partitions with width converging to 0. Then Spn (f ) converges
to f .
Proof. Given ε > 0, and let δ > 0 be the magic number for
ε
b−a
in f . The claim is that if a mesh p has width less than δ then Sp (f ) is within ε of
Check this.
26
R
f.
Evan Chen (Spring 2015)
6 February 19, 2015
§6 February 19, 2015
Didn’t attend lecture for HMMT reasons; however the lecture was just a review of
homework questions. The notes below are due to W Mackey.
§6.1 Bonus Problem Sets
The first bonus PSet on normed vector spaces has been posted. The plan for the bonus
PSets hereon will be:
1. Banach Spaces
2. Topological vector spaces
3. Operator Theory
4. Spectral Theory in Hilbert Spaces
5. Lebesgue measure
6. Lebesgue Integration
Today we’ll mostly be doing a ‘problem solving exercise’ rather than any new material.
§6.2 PSet 2, Problem 7b
Let X be a metric space; let Cauch(X) and let X be the quotient of Cauch(X) by
the equivalence relation that {xn } ∼ {yn } if and only if ρ({xn }, {yn }). Show that X
is complete in this metric.
Let n be fixed and yn = {xm
n } be a Cauchy sequence of elements indexed by m. We
want to show that yn → y. Define a sequence by fixing a sequence of real numbers
rn → 0. Let y = {xn }, and we want to construct the sequence xn . Pick N so that for
n1 , n2 ≥ N , ρ(yn1 , yn2 ) < 13 rn . Pick any n0 ≥ N , and consider yn0 = {xm
n0 }. Let M be so
0
m1
m2
1
0
that m1 , m2 ≥ m implies ρ(xn0 , xn0 ) < 3 rn . Pick m ≥ M . Set xn := xm
n0 . Then we
claim this is Cauchy. For any , if rN < then n1 , n2 ≥ N , ρ(xn1 , xn2 ) < .
Alternate method: Suppose we prove X is dense. Then X ,→ X.
Lemma 6.1
X is dense in X.
Assume for the moment the lemma. Then take {xn } a Cauchy sequence. Then pick
xn ∈ X such that ρ(xn , xn ) < n1 , then this sequence will converge as desired.
The lemma is almost definitional.
§6.3 PSet 3, Problem 8
Let X and Y be topological spaces with Y compact. Show that the projection
X × Y → X sends closed subsets to closed subsets.
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Evan Chen (Spring 2015)
6 February 19, 2015
We want P (Z) to be closed for every Z. We’ll prove this on the opens instead. Let
x ∈ X, x 6∈ P (Z). We want x ∈ U ⊆ X such that U ∩ P (Z) = ∅. For all y ∈ Y , we have
(x, y) 6∈ Z. Then there exist Ux , Uy such that (x, y) ∈ Ux × UY and Ux × Uy ⊆ X × Y − Z.
Now take an open cover ∪Uy = Y for every point, then take a finite subcover ∪Uyi = Y .
Then fix U = ∩i UYi , and we claim U ∩ P (Z) = ∅, proving the complement is empty,
hence the projection of Z is closed.
§6.4 PSet 3, Problem 10
Consider the set X = (0, 1) ∪ ∞. Define a topology on it, so that in addition to
X and ∅, the open subsets are: (i) any U which is open in (0, 1); (ii) the subsets
∞ ∪ (0, a) ∪ (b, 1) for 0 < a < b < 1; (iii) unions of the above. Problem 10b was
showing that X is compact.
Let X be a topological space such that for every x ∈ X and every open x ∈ Ux open,
there exists a x ∈ Ux0 ⊆ Yx ⊆ Ux where Ux0 is open and Yx is compact. We call this
space locally compact. Let X = X ∪ {∞}, with X Hausdorff. A set in X is open if
either it’s empty or the entire thing, it’s contained X and open there, or its complement
is a compact subset of X. Note that in question 10, this is exactly the topology we
described.
Lemma 6.2
X is compact.
Proof. Say X = ∪α Uα . There exists an α0 such that ∞ ∈ Uα0 since it’s a cover. Consider
Z = X − Uα0 , is compact by definition. Then Z ⊆ Uα (Uα − ∞), then compactness of Z
means we can take finitely many there, so the adjoining Uα0 gives us our finite cover.
Now we just want to verify that X is a topology. Take U open and Z compact, then
((X − Z) ∪ {∞}) ∪ U is open, (X − U ) ∩ Z is closed in Z and therefore still compact.
Now for finite intersections, take U ∩ ((X − Z) ∪ {∞}) = (X − Z) ∩ U is open.
Lemma 6.3
X is Hausdorff if and only if X is locally compact.
Proof. Fix x ∈ X, ∞ ∈ X. Hausdorff means there is a U ⊆ X open and an open set
that contains ∞ and not x such that the two are disjoint, so there is compact Z ⊆ X
such that U ∩ (X − Z) = ∅, implying U ⊆ Z. (TL;DR Hausdorff gives us an open set
that hides in the complement, which is compact definitionally.)
§6.5 PSet 3, Problem 4
(a) Let r be a real number Show that for x ∈ R>0 for any sequence of rationals
ri → r, the sequence xri is Cauchy. We set xr to be its limit.
Write xrm − xrn = xrn (xrm −rn − 1). Let’s assume x ≥ 1, we’ll get x < 1 by inversion.
Then we want to say x1/n − 1 ≤ n1 (x − 1). Our first lemma will be xr for r in Q is
monotonic. Our second will be that x1/n → 1. This will work out.
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Evan Chen (Spring 2015)
6 February 19, 2015
(b) Prove that the function x 7→ xr : R>0 → R>0 is continuous and satisfies
(x1 · x2 )r = (x1 )r · (x2 )r .
We know that xri − xr = xr
xi r
x
r
− 1 , and xxi is monotonic in r. Then
x r
x r0
i
i
− 1 ≤ − 1 .
x
x
by dominating with a rational number. This sandwiches
x R
i
x
≥
x r
i
x
≥1
for xi ≥ x, and the other way around for xi ≤ x, then we squeeze.
(c) Prove that for a fixed a ∈ R>0 , the function x 7→ ax : R → R>0 is continuous and
satisfies
ax1 +x2 = ax1 · ax2 .
Another squeeze argument. We have ax (ax−xi − 1), and ri0 ≥ x − xi ≥ ri , we know
→ 1 for ri rationals that tend to 0, so we get what we want.
ari
(d) Assume that a 6= 1. Show that for y ∈ R>0 there exists an element loga (y) ∈ R
so that aloga (y) = y, and that the resulting function y 7→ loga (y) : R>0 → R is
continuous.
This is just since the inverse function is also continuous here, which we’ve proved
before.
Lemma 6.4
f : [a, b] → R is continuous and strictly monotonic on the rationals. Then it is
monotonic.
Proof. Take x < y, let x ← xi and yi → y, then f (x) = lim f (xi ) < lim f (yi ) = f (y).
Then we get that ax is surjective by the intermediate value theorem.
§6.6 Filters
(Something Gaitsgory specifically didn’t want to do) You can test limits in arbitrary
topological spaces with a filter. Sequences aren’t always enough. To test convergence in
a topological space that isn’t first countable, you have to map other types of ordered sets.
Gaitsgory says that he doesn’t find it very useful. It’s much better to check compactness
and such by the general topological, rather than sequential definitions. For more, see:
http://en.wikipedia.org/wiki/Filter_%28mathematics%29.
§6.7 PSet 4
PSet 4 is mostly functional analysis stuff. For problem 1, Functcont (X, Y ), for X compact
and Y metric (complete). We’ll say ρ(f, g) = maxx∈X ρ(f (x), g(x)). The theorem is that
the space of functions is complete.
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Evan Chen (Spring 2015)
6 February 19, 2015
For problem 2, we have Y = R, and ρ(f, g) = ρ(f −h, g−h), further ρ(cf, cg) = |c|ρ(f, g).
Problem 2 is pretty boring.
Problem 3 is interesting. We say two norms are equivalent if there exist scalars A, B
such that the first norm is bounded by the second up to A, and the second is bounded
by the first up to B. Problem 3 is proving that any two norms are equivalent in finite
dimensional vector spaces.
Problem 4 shows this is emphatically not the case for infinite dimensional vectorR spaces.
In particular, we compare the max norm and the L1 norm, defined by ||f ||L1 = [a,b] |f |.
We’ll show that these two norms are not equivalent.
Problem 5 introduces the L2 norm and is fairly easy.
Problem 6 is showing a way to prove second countability.
Problem 7 is also kind of cool. We consider the completion of Functcont (X, Y ) with
the L1 norm, L1 ([a, b]). We then do it for L2 ([a, b]).
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Evan Chen (Spring 2015)
7 February 24, 2014
§7 February 24, 2014
Let f : X → Y be a map of topological spaces.
§7.1 Limits
Definition 7.1. We say f is continuous at x if for all neighborhoods Uy 3 f x there
exists a neighborhood Ux 3 x then
f “(Ux ) ⊆ Uy .
Definition 7.2. Let f˚ : X − {x} → Y . We say that the limit of f˚ at x is y if f˚ can be
extended to f : X → Y so that f (x) = y and f is continuous at x.
Equivalently, the limit of f at x is y if for every neighborhood Uy 3 y there exists a
neighborhood Ux 3 x such that ∀x0 ∈ Ux − {x0 } we have f (x0 ) ∈ Uy .
We as usual write
lim f (x) = L
x→c
to mean that L is the limit of f at c.
Remark 7.3. Given Y = R, limits are additive. Moreover, for any bounded function
g : X → R, if limx→c f (x) = 0 then limx→c f (x)g(x) = 0.
Assume now that X is a metric space. We introduce little o notation:
Definition 7.4. A function f : X − {x0 } → R is o(ρ(x, x0 )n ) if
lim
x→x0
f (x)
= 0.
ρ(x, x0 )n
Equivalently, ∀ε > 0, ∃δ > 0 such that
ρ(x, x0 ) < δ =⇒ |f (x)| ≤ ε |ρ(x, x0 )|n .
We restrict our attention to nonnegative integers n.
In particular, f is o(1) if limx→x0 f (x) = 0. (This is n = 0 above.)
Example 7.5
Let X = R, f (x) = x3 + x sin x. Then f (x) is o(xn ) for n = 0, 1, 2, 3 only.
§7.2 Differentiation
Consider a function f : (a, b) → R.
Definition 7.6. We say f is differentiable at x0 ∈ (a, b) if there exists a constant C
such that
f (x) − f (x0 ) − (x − x0 )C = o(x − x0 )
id est,
lim
x→x0
f (x) − f (x0 ) − C(x − x0 )
f (x) − f (x − 0)
= 0 ⇐⇒ lim
= C.
x→x0
x − x0
x − x0
In that case we write f 0 (x0 ) for the constant C.
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Evan Chen (Spring 2015)
7 February 24, 2014
Lemma 7.7
If f 0 (x0 ) exists then it is unique, and f is continuous at x0 .
Proof. For the first part, limits are unique. The second part is some boring calculation
(the point is that differentiability is stronger than continuity).
“If you find this exceedingly boring, quietly go to sleep for the next ten
minutes. Actually, maybe all of today will be boring. Go check your Facebook
or something.”
– Gaitsgory
Example 7.8
If f (x) = x then f 0 (x0 ) = 1 for any x0 .
Lemma 7.9 (Product Rule)
If f and g are differentiable at x then so is f · g and the value is given by
(f · g)0 (x) = f 0 (x)g(x) + f (x)g 0 (x0 ).
Proof. Bash.
As usual, we can by induction get the derivative for xn . It doesn’t work for x1/n because
you don’t know a priori that x1/n is differentiable.
§7.3 Chain Rule
Lemma 7.10 (Chain Rule)
f
g
Let (a, b) −
→ (c, d) →
− R. Suppose f is differentiable at x0 and g is differentiable at
y0 = f (x0 ). Then (g ◦ f )0 (x0 ) exists and has value
g 0 (y0 ) · f 0 (x0 ).
Proof. Do a few tricks:
gf x − gf x0 − g 0 (y0 )f 0 (x0 )(x − x0 )
x→x0
x − x0
gf x − gy0 − g 0 y0 (f x − y0 )
f x − y0 − (x − x0 )f 0 x0
=
+ g 0 y0 ·
x − x0
x − x0
0
gf x − gy0 − g y0 (f x − y0 )
=
x − x0
lim
Thus this amounts to showing that
gf x − gy0 − g 0 y0 (f x − y0 )
→ 0.
x − x0
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Evan Chen (Spring 2015)
7 February 24, 2014
For ε > 0 there’s a µ > 0 such that for |y − y0 | < µ we have
gy − gy0 − g 0 y0 (y − y0 ) <ε
y − y0
and take δ > 0 such that
|x − x0 | < δ =⇒ |f x − y0 | < µ
so
gy − gy0 − g 0 y0 (y − y0 ) ≤ ε · f (x) − y0 .
y − y0
x − x0
§7.4 More “applied math”
Theorem 7.11 (Critical Points)
Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Then if y
attains a maximum at x0 ∈ (a, b) then f 0 (x0 ) = 0.
Proof. Assume not, and WLOG f 0 (x0 ) > 0. Take a small perturbation.
Theorem 7.12 (Rolle)
Let f : [a, b] → R be continuous and differentiable on (a, b). Then for some x ∈ (a, b)
we have f 0 (x) = 0.
Proof. Compactness to get maximum.
“Imagine you’re going on I-90, going east until at some point you stopped
and turned back because Massachusetts is the best.”
By previous theorem, f 0 (x) = 0.
But what if x = a?
“So imagine you didn’t travel east, you travelled west instead.”
Take a minimum instead.
What if the minimum and maximum are both endpoints?
“If the maximum is at the endpoint and the minimum is at the endpoint,
then you really have a problem; the function is constant so you didn’t go
anywhere. No place like home.”
Theorem 7.13 (Mean Value Theorem)
With the same assumptions, there exists x ∈ (a, b) such that
f 0 (x) =
f (b) − f (a)
.
b−a
Proof. By Gaitsgory quote.
“Suppose you’re going at I-90. In an hour you cover 95 miles. The cop claims
at some point you were travelling at least 95 miles per hour.”
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Evan Chen (Spring 2015)
7 February 24, 2014
Theorem 7.14
Let f be differentiable on (a, b).
(a) If f 0 (x) ≥ 0 for all x then f is nondecreasing.
(b) If f 0 (x) ≤ 0 for all x then f is nonincreasing.
(c) If f 0 (x) = 0 for all x then f is constant.
Proof. Use MVT for (a) and (b); then (c) follows.
§7.5 Higher Derivatives
Definition 7.15. Let f : (a, b) → R. We say f is differentiable n times at x0 if it’s
differentiable n − 1 times on (a0 , b0 ) and then f (n−1) is differentiable at x0 .
Theorem 7.16 (Stop Sign Theorem)
Let f be differentiable n times at x0 such that
f (n) (x0 ) = 0 k = 0, . . . , n.
Then
f = o ((x − x0 )n ) .
Proof. Induction on n.
Theorem 7.17 (Taylor)
Let f be differentiable n times at x0 . Then
f (x) −
n
X
f (i) (x0 )(x − x0 )i
i!
i=0
is o ((x − x0 )n ).
Proof. Apply previous theorem.
Unfortunately, even “infinitely differentiable with all zero derivatives” is not enough to
guarantee that f is identically zero in any neighborhood.
Proposition 7.18
The function f : R → R by
(
0
x 7→
e−1/x
x≤0
x > 0.
is infinitely differentiable at x = 0, with all derivatives zero.
Proof. The point is to check that
e−1/x
xn
→ 0 as x → 0 for every n.
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Evan Chen (Spring 2015)
8 February 26, 2015
§8 February 26, 2015
Today we prove the Fundamental Theorem of Calculus.
Lemma 8.1
A function [a, b] → R whose derivative on (a, b) is everywhere zero is constant.
Proof. Mean Value Theorem.
§8.1 Fundamental Theorem of Calculus
Let f : [a, b] → R be continuous. We define F : (a, b) → R by
Z
f.
F (x) =
[a,x]
Theorem 8.2
F is differentiable and
F 0 (x) = f (x).
Proof. It suffices to show that
R
[x,x0 ] f
x →x x0 − x
lim
0
→ f (x)
which is not difficult (just ε and δ using continuity of f , and bound the integrand).
Corollary 8.3 (Fundamental Theorem of Calculus)
Let F be continuous on [a, b] and differentiable on (a, b) such that F 0 extends to a
continuous function on [a, b]. Then
Z
F (b) − F (a) =
F 0.
[a,b]
Rx
Proof. Let G(x) = a F 0 . By the previous theorem, G is differentiable on (a, b) and
G0 = F 0 . You can check it’s also continuous at x = a and x = b. Since G − F has zero
derivative, it is constant. Thus
Z
G(b) − F (b) = G(a) − F (a) = −F (a) =⇒ F (b) − F (a) = G(b) =
F 0.
[a,b]
You can extend this all to work for certain noncontinuous functions, but
“Trying to Riemann integrate discontinuous functions is kind of outdated.”
– Gaitsgory
We want Lebesgue integrals.
Remark 8.4. As we’ve said many times, and will continue to see: integrals are much
nicer than derivatives. It’s easy to control the integral of continuous functions compared
to controlling derivatives (or even proving derivatives exist).
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Evan Chen (Spring 2015)
8 February 26, 2015
§8.2 Pointwise converge sucks
Recall that
X xn
exp(x) =
n≥0
n!
.
We wish to show exp0 = exp. In an ideal world, we would have the following result:
Let fn be a sequence of functions [a, b] → R differentiable on (a, b) such
that for every x, the sequence f1 (x), f2 (x), . . . converges to f x. Then f is
differentiable and
f 0 (x) = lim fn0 (x).
n→inf ty
Unfortunately this notion of “point-wise convergence” is not good enough, which is to
say that the theorem is false. Here’s a counterexample. We define a sequence gn by


0 ≤ x ≤ 21 − n1
0
1
1
1
1
gn (x) = 12 n x − ( 12 − n1 )
2 − n ≤x≤ 2 + n


1
1
1
2 + n ≤x≤1
Rx
and let fn (x) = 0 gn (t) dt. It is not difficult to see that
(
0
x ≤ 12
lim fn = x 7→
n→∞
x − 21 x ≥ 12 .
§8.3 Uniform Convergence of Functions
Definition 8.5. Let f1 , . . . be a sequence of functions X → R. We say that (fn )
converges uniformly to f : X → R if they converge in the sup metric, id est
lim sup |f (x) − fn (x)| = 0.
n→∞ x∈X
In other words, for every ε > 0 the quantity
|f (x) − fn (x)|
is bounded by ε for all sufficiently large n.
The less happy notion of just
lim |f (x) − fn (x)| = 0
n→∞
is the one we tried above.
Here is another false theorem.
Theorem 8.6
Let f be a sequence of continuous functions on [a, b] differentiable on (a, b). Assume
that each fn0 extends to a continuous function on [a, b], and moreover the sequence
f10 , f20 , . . . converges uniformly to g. Assume also that for some c ∈ [a, b] the sequence
f1 (c), f2 (c), . . . converges.
Then fn converges uniformly to some differentiable function f whose derivative
equals g.
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Evan Chen (Spring 2015)
8 February 26, 2015
Proof. Let
Z
f (x) =
[a,x]
where C =
limn→∞ fn (c) −
R
[a,c] g
g−C
(the constant C is contrived so that f (c) =
limn→∞ fn (c)). By the Fundamental Theorem of Calculus, f 0 = g. So now we simply want to show fn → f .
We write f and fn so they look similar:
!
Z
Z
fn (x) =
[a,x]
fn0 +
fn (c) −
Z
f (x) =
[a,c]
lim fn (c) −
g+
n→∞
[a,x]
fn0
!
Z
g
[a,c]
We claim that each term converges uniformly to the corresponding term. That is, we
have fn (c) → limn→∞ fn (c) by definition. Given ε → 0 have for suitably large n that
Z
Z
0
fn − g <
ε→0
[a,x]
and similarly for the
0
[a,c] fn
R
[a,x]
term.
§8.4 Applied Math – Differentiating ex
Now we finally prove the following result.
Theorem 8.7
The function exp : R → R is differentiable and equals its own derivative.
Proof. Restrict to any interval [a, b]. Since
exp(x) =
X xn
n≥0
we define
fn (x) =
n!
n
X
xk
k=0
k!
.
It is pretty amusing to show that fn0 = fn−1 . Hence, fn → exp(x) and fn0 → exp(x) as
well. Thus we only need to show that the convergence is uniform, and we’ll be done.
For any n we have
k
X (max {|a| , |b|})k n→∞
X
x ≤
|exp(x) − fn (x)| = −→ 0.
k!
k≥n+1 k! k≥n+1
Also 1 = f1 (0) = f2 (0) = . . . . So the theorem applies and we’re done.
Second Proof. We can do this directly using the fact that we proved exp(x + y) =
exp(x) + exp(y). Observe that
ex+h − ex
eh − 1
= ex ·
.
h
h
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Evan Chen (Spring 2015)
8 February 26, 2015
So we merely need to show that limh→0
lim
h→0
eh −1
h
= 1, which is
X hk−1
k!
k≥1
= 1.
The idea is to bound k! by a geometric series now:
1
1
X hk−1
X hk−1
2 h 1
2
− 1 = h + ( h) + . . . = − 1 ≤ → 0.
k−1
1 − 21 h 2
2
k≥1 k!
k≥1 2
§8.5 Differentiable Paths in Rn
Consider a function γ : (a, b) → Rn .
Definition 8.8. We say γ is differentiable if each of the coordinates is also differentiable.
Let γ : [a, b] → Rn be a continuous function differentiable on (a, b). We wish to define
a notion of length on γ.
Definition 8.9. Given a norm k•k on Rn we define the length by
Z
0
γ .
length(γ) =
[a,b]
A mesh of points p is, as before, a sequence
a = c0 ≤ c1 ≤ · · · ≤ cm = b.
The width is maxk ck+1 − ck . Then
Theorem 8.10
Given any sequence of partitions pn with width approaching zero, we have that


X
length(γ) = lim 
kγ(ck+1 ) − γ(ck )k .
n→∞
[ck ,ck+1 ] in pn
Proof. Homework, but via Riemann sums the problem boils down to showing that


X
(ck+1 − ck )γ 0 (di ) .
lim 
n→∞
di ∈[ck ,ck+1 ] in pn
converges to the same thing as

lim 
n→∞

X
[ck ,ck+1 ] in pn
kγ(ck+1 ) − γ(ck )k .
For this we will need to show that
Z
Z
f ≤
kf k .
[a,b] [a,b]
38
Evan Chen (Spring 2015)
9 March 3, 2015
§9 March 3, 2015
Guess who showed up half an hour late to class? Blocking is hard.
§9.1 Differentiable forms
In what follows, V and W are real finite-dimensional vector spaces and U is an open
subset of V .
Definition 9.1. Let V and W be normed vector spaces. We say f : U → W is
differentiable at x (here x ∈ U ) if there exists a function T ∈ HomR (V, W ) such that
kf (x + v) − f (x) − T (v)kW
= 0.
v→0
kvkV
lim
We write Df (x) for the linear transformation T .
Here Df (x) is a linear map of vector spaces. If W = R, we find Hom(V, R) = V ∨ is
the dual module. When V = R, V ∨ consists of scalar functions and this looks like our
usual notion of derivative.
The following definitions are my best guesses, since I was very late to lecture.
Definition 9.2. Let x ∈ U and v ∈ V . For f : U → W differentiable we write
(Dv f )(x) = (Df (x))(v).
Hence Dv f : V → W .
Note that Dcv (f ) = cDv f for any constant c.
Definition 9.3. Let f : Rn → R be differentiable and let e1 , . . . , en be a distinguished
basis. Then we define ∂i f as shorthand Dei (f ).
Lemma 9.4
A function f : U → W which is differentiable at x is also continuous at x.
Proof. Same as one variable case.
Theorem 9.5
Assume that the partials ∂i f exist and are continuous on some x ∈ U 0 ⊆ U . Then f
is differentiable.
Proof. Note that
f (x, y) − f (0, 0) − x∂x f (0, 0) − y∂y (0, 0)
p
x2 + y 2
equals
f (x, y) − f (0, y) − x∂x f (0, 0) f (x, y) − f (0, ) − y∂y f (0, 0)
p
p
+
x2 + y 2
x2 + y 2
The first part equals
f (x, y) − f (0, y) − x∂x f (0, 0)
p
≤ ∂x f (x0 , y) − ∂x f (0, 0) → 0.
x2 + y 2
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Evan Chen (Spring 2015)
9 March 3, 2015
§9.2 Vector-Valued Functions
f
Let U1 −
→ U2 , where U1 ⊆ V1 and U2 ⊆ V2 are open.
Lemma 9.6
Let f = (f1 , . . . , fm ) be a function V1 → Rm . Then f is differentiable if and only if
each fi is differentiable and
πi (Df (x)) = Dfi (x).
Theorem 9.7 (Chain Rule)
f
g
Consider U1 −
→ U2 →
− U3 , where Ui is open in a real vector space Vi . Suppose f is
differentiable at x1 ∈ U1 and g is differentiable at x2 = f (y1 ) ∈ U2 . Then
D(g ◦ f )(x1 ) = Dg(x2 ) ◦ Df (x1 ).
Proof. Same as one-variable case.
§9.3 Higher-Order Derivatives
Definition 9.8. Let f : U 0 → W , where V and W are vector spaces (here U 0 ⊆ V is
open). Then f is differentiable twice at x if
(1) f is differentiable on some neighborhood U of x, so we may consider Df .
(2) The map Df : U → Hom(V, W ) is itself differentiable at x.
For three times, we want
D2 f : U → Hom(V, Hom(V, W )) ∼
= Hom(V ⊗2 , W )
to itself be differentiable at x.
Theorem 9.9
Let f : V → W be twice differentiable. Let v, w ∈ V and consider Dw (Dv (f )) and
Dv (Dw (f )). If both are continuous then they coincide.
In particular,
Corollary 9.10
“Let f be diff be differentiable twice” (sic). Assume that ∂j (∂i f ) and ∂i (∂j f ) are
continuous. Then they coincide.
In fact this is the only case we need consider, since if v and w are linearly dependent it’s
trivial.
Proof. Assume V = R2 , v = e1 , w = e2 , since the case where v and w are linearly
dependent is trivial.
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Evan Chen (Spring 2015)
9 March 3, 2015
We may assume by shifting and subtracting linear factors that x = (0, 0), f (0, 0) = 0
and Df (0, 0) = 0.
Define
f (x, y) − f (x, 0) − f (0, y) + f (0, 0)
H(x, y) =
.
xy
We claim that
lim H(x, y) = ∂y ∂x f.
x→0
y→0
The claim will then follow by symmetry.
−f (0, y)
∂y
+f (x, y)
ỹ
+f (0, 0)
∂x
−f (x, 0)
x̃
By using Mean Value Theorem twice as in the diagram above, we can do a trick where
we force
H(x, y) = ∂y ∂x (x̃, ỹ)
where 0 ≤ x̃ ≤ x, 0 ≤ ỹ ≤ y. Since x, y → 0 we get the desired lemma.
41
Evan Chen (Spring 2015)
10 March 5, 2015
§10 March 5, 2015
§10.1 Inverse function theorem
Let f : U1 → U2 be continuously differentiable, where U1 ⊆ V1 and U2 ⊆ V2 are open.
We’re interested in whether there exists a g such that g is the inverse of f .
Suppose for now g does exist, so that we have
f
V1 ⊇ U1 - U2 ⊆ V2
g
We have f ◦ g = id1 and g ◦ f = id2 , so given x1 ∈ U1 and x2 = f (x1 ) ∈ U2 we obtain
Df (x1 ) : V1 → V2
and Dg(x2 ) : V2 → V1
The chain rule implies that
Df (x1 ) ◦ Dg(x2 ) = id2
and Dg(x2 ) ◦ Df (x1 ) = id1 .
Thus it’s necessary that Df (x1 ) be invertible.
Remark 10.1. Remember that Df (x1 ) is a linear map. Hence the condition that Df (x1 )
is invertible is not terribly impressive; it’s equivalent to some determinant not being zero.
The inverse function theorem states that this weak condition is in factlocally sufficient.
Theorem 10.2 (Inverse Function Theorem)
Let U1 ⊆ V1 and U2 ⊆ V2 be vector spaces, and let f : U1 → U2 be continuously
differentiable. Suppose Df (x1 ) is invertible for some x1 ∈ U1 (in particular dim V1 =
dim V2 ). Then there exists U10 3 x1 and U20 3 x2 = f (x1 ) neighborhoods such that
the restriction f : U10 → U20 admits a differentiable inverse.
Remark 10.3. It turns out that in order to get a global inverse we require U2 to be
simply connected.
Example 10.4
For example, if f : R → R the fact that Df x is invertible is just the condition that
f 0 (x1 ) 6= 0. So a function f : R → R is locally invertible if and only if the derivative
at that point is not zero.
Example 10.5
Needless to say, global inverses may not exist. For example, consider C 7→ C \ {0}
(where C ' R2 ) by
z 7→ exp(2πiz).
Every neighborhood is invertible but the target space is globally not.
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Evan Chen (Spring 2015)
10 March 5, 2015
§10.2 Proof of the Inverse Function Theorem
The proof follows the following three steps.
1. We show that it’s possible to restrict to a small enough neighborhood so that f is
injective.
2. We show that as long as Df is invertible, the function is open, meaning it maps
open sets to open sets. Hence we can let g = f −1 as a function of sets; the fact
that f is open then yields continuity of g.
3. Finally, we show that g is differentiable.
The hard part is the second step.
Let U = U1 . Throughout the proof, we will continuously shrink U and let U shrink;
we’ll continue using the same letter U .
§10.3 Step 0
Consider norms k−k1 and k−k2 on V1 and V2 . Recall (from the homework) that for a
linear map T : V1 → V2 we defined the norm of a linear operator is given by
def
kT kop = max kT (v1 )k2 .
kv1 k1 ≤1
This endows HomR (V1 , V2 ) with a metric topology.
Lemma 10.6
Let HomR (V1 , V2 )inv denote the set of invertible linear maps. Then it is open in
HomR (V1 , V2 ).
Proof. It’s vacuous if dim V1 6= dim V2 since HomR (V1 , V2 )inv is empty in that case.
Now recall that amp is invertible if and only if it has zero determinant, and
det : Matn×n (R) → R
is continuous because “it consists of multiplying things”.
“This is classic Math 55, we spend a whole semester learning abstract linear
algebra and then we’re here and no one knows what to do.” – James Tao
Lemma 10.7
The map from HomR (V1 , V2 )inv to itself by T 7→ T −1 is continuous.
Proof. Use applied math – there is a formula for T −1 in terms of multiplying blah.
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Evan Chen (Spring 2015)
10 March 5, 2015
Corollary 10.8
IF Df (x1 ) is invertible then there is a neighborhood U ⊆ U1 such that Df (x−1
1 ) is
invertible for all x ∈ U and the map
x 7→ (Df (x))−1
is continuous.
§10.4 First Step
Proposition 10.9
Let U ⊆ U1 be convex, and let (Df (x1 ))−1 op ≤ Λ over f : U → U2 . Moreover,
assume Df is uniformly continuous. Then for some δ > 0, if kx01 − x001 k < δ then
f (x01 ) − f (x001 ) ≥ 1 x01 − x001 .
1
2
2Λ
This will imply Step 1, because by taking r small enough we can consider a closed r-ball
B of x1 on which Df (x01 ) is invertible. This
closed
ball is compact, so Df continuous
−1
implies Df is uniformly continuous; thus (Df ) op is bounded on B. Applying the
proposition, we let U be an open ball of radius less than half the specified δ, contained
entirely inside B. Then the proposition implies the conclusion.
To prove Proposition 10.9 we first show the following Proposition 10.10.
Proposition 10.10 (“Uniform” Differentiability)
Let g : U1 → U2 be such that Dg is uniformly continuous, where U1 is convex. Then
for all ε > 0 there exists δ > 0 such that whenever kx0 − x00 k1 < δ we have
kg(x00 ) − g(x0 ) − Dg(x0 )(x00 − x0 )k2
< ε.
kx00 − x0 k1
Proof. Identify V2 = Rn with the max norm. In this way it suffices to consider V2 = R
(i.e. we consider everything componentwise).
This allows us to appeal to the Mean Value Theorem. Thus we find that there exists x̃
lying on the line segment joining x0 to x00 for which
Dg(x̃)(x0 − x00 ) = g(x0 ) − g(x00 ).
(Here we use convexity so that Dg is defined on the entire segment from x0 to x00 .) Thus
the fraction is question is bounded by the operator norm
Dg(x̃) − Dg(x0 ) .
op
Hence the conclusion follows by uniform continuity.
Proof of Proposition 10.9 from Proposition 10.10. Let ε =
sition 10.10.
44
1
2Λ
and let δ be as in Propo-
Evan Chen (Spring 2015)
10 March 5, 2015
Define
v = f (x001 ) − f (x01 ) − Df (x01 )(x001 − x01 ).
We are given that
kvk2 <
1 x00 − x0 .
1
2Λ
Apply (Df (x01 ))−1 to v. On one hand, we have
(Df (x01 ))−1 (v) ≤
1
x001 − x01 .
1
2
On the other hand
(Df (x01 ))−1 (v) = (Df (x01 )−1 )(f (x001 ) − f (x01 )) − (x001 − x01 )1 .
By Triangle Inequality, we thus derive
(Df (x01 )−1 )(f (x001 ) − f (x01 )) ≥ 1 x001 − x01 .
1
2
Hence
f (x001 ) − f (x01 ) ≥ 1 x001 − x01 .
2
2Λ
§10.5 Third Step
In the spirit of procrastination, we now skip to step three.
Let f : U10 → U20 be a injective and surjective. Assume that Df is invertible and open,
so that it has a continuous inverse g. We wish to show that g is in fact continuous.
We claim that if f (x1 ) = x2 , then
Dg(x2 ) = Df (x1 )−1
(indeed, we know a priori that this ought to be the case).
We wish to show now that
0
g(x ) − g(x2 ) − (Df (x1 )−1 )(x0 − x2 )
2
2
1
lim
= 0.
kx02 − x2 k2
x02 →x2
Let x2 = f (x1 ) and x02 = f (x01 ). So the expression rewrites as
Df (x1 )−1 (Df (x1 )(x0 − x1 ) − (f (x0 ) − f (x1 )))
1
1
kf (x01 ) − f (x1 )k2
1
.
This is at most
0
0
kx01 − x1 k
Df (x1 )−1 · kf (x1 ) − f (x1 ) − Df (x1 )(x1 − x1 )k ·
.
0
kx1 − x1 k
kf (x01 ) − f (x1 )k
The first term is bounded, the middle term tends to zero (by definition), and the last
term is bounded by some constant due to Proposition 10.9.
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Evan Chen (Spring 2015)
10 March 5, 2015
§10.6 Second Step
We do so in complete generality.
Proposition 10.11
Let f : U1 → U2 be any continuous function (not necessarily injective), so that
Df (x1 ) is invertible at each x1 ∈ U1 .
Then for all x1 ∈ U1 , there exists ε > 0 and r > 0 such that
B2 (f (x1 ), ε) ⊆ f “(B1 (x, r)).
(Here as usual, f “(S) = {f (s) | s ∈ S} is the pointwise image, while Bi denotes open balls
in Vi ). This is enough to show that f “(open) is open (the balls form a basis), meaning
that f is an open mapping.
(Note: the “U1 ” here is not the same as for our original function Specifically, to show
that g : U1 → U2 is an open mapping we are going to apply the proposition g|U for some
U ⊆ U1 .)
First, we prove the following theorem on contractions, which we’ll continue to use in
the next lecture repeatedly!
Theorem 10.12 (Contractions of Complete Spaces Have Unique Fixed Points)
Let X be a complete metric space with metric ρ, and let φ : X → X a contraction,
meaning for some 0 < λ < 1 such that
ρ(φ(x1 ), φ(x2 )) ≤ λρ(x1 , x2 ).
Then φ has a unique fixed point.
“There is a traffic interpretation of this. The gas tank of your car has a leak
so eventually you run out of gas.” – James Tao
Proof. Uniqueness is straightforward; if φ(x1 ) = x1 and φ(x2 ) = x2 then
ρ(x1 , x2 ) < λφ(x1 , x2 ).
For existence, consider the Cauchy sequence
x φ(x) φ(x2 ) . . . .
By completeness, it converges to some point p. You can check that φ(p) = p.
To be finished next week. . .
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Evan Chen (Spring 2015)
11 March 10, 2015
§11 March 10, 2015
§11.1 Completing the Proof of Inverse Function Theorem
Definition 11.1. A function f : X1 → X2 is open if it sends open sets in X1 to open
sets in X2 .
This is the reverse of our typical continuity in which we require the pre-image of an
open set in X2 to be an open set in X1 . In the same way that this gives an ε-δ continuity
for metric spaces, we can rewrite this as
(∀x1 ∈ X1 )(∀r > 0)(∃ε > 0)f “ (B(x, r)) ⊇ B (f x1 , ε) .
By fiddling with the radius, we can replace this with
(∀x1 ∈ X1 )(∀r > 0)(∃ε > 0)f “ B(x, r) ⊇ B (f x1 , ε) .
where B is the closed a ball of radius r.
We need only prove the following result now.
Theorem 11.2
If Df (x1 ) is invertible, then f is open on a neighborhood of x1 .
Proof of Theorem. Let ε > 0 be small enough to be determined later, and let x02 ∈
B(f x1 , ε) be arbitrary. We wish to show there exists an x01 so that f x01 = x02 .
We define a map Φ : B(x1 , r) → V1 by
Φ(x01 ) = x01 − (Df x1 )−1 f (x01 ) − x02
Observe that finding x01 as above amounts to finding a fixed point of Φ.
We wish to shrink r sufficiently so that
Φ(x001 ) − Φ(x01 ) ≤ 1 x001 − x01 .
2
Compute
Dφ(x01 ) = (Df (x1 ))−1 Df (x1 ) − Df (x01 ) .
Let Λ = Df f (x1 )−1 , so that
Dφ(x001 ) ≤ Λ Df (x1 ) − Df (x01 ) .
We can choose r small enough that the right-hand side is at most 12 , by forcing
1
kDf (x1 ) − Df (x01 )k ≤ 2Λ
(continuous functions can have arbitrarily small difference).
Recall from a previous problem set that said that We now invoke the following result
from Problem Set 6.
1(c). Let U ⊂ V be convex and let f : U → W be a continuously differentiable
function. Let W be also endowed with a norm, and suppose that kDf (x)k ≤ Λ for
all x ∈ U . Show that
f (x0 ) − f (x00 ) ≤ Λ · x0 − x00 , ∀x0 , x00 ∈ U.
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Evan Chen (Spring 2015)
11 March 10, 2015
Applying the result to g = Dφ gives the desired estimate, so that Φ is a contraction by a
factor of 12 .
It remains to show that Φ maps B(x1 , r) into itself, at which point we can apply the
fact that contractions have fixed points. We have that
Φ(x01 ) − x1 = Φ(x01 ) − Φ(x1 ) + kΦ(x1 ) − xk
1
≤ x01 − x1 + kΦ(x1 ) − x1 k
2
1
≤ r + kΦ(x1 ) − x1 k
2
1
≤ r + (Df x1 )−1 (x2 − x02 )
2
1
≤ r + (Df x1 )−1 (x2 − x02 )
2
1
= r + Λ (x2 − x02 ) .
2
If we select the ε earlier to guarantee that kx2 − x02 k <
r
2Λ
then we’re done.
§11.2 Implicit Function Theorem
Here is the “high-school version” Suppose f : R2 → R and f (x0 , y0 ) = z0 , and moreover
∂x f (x0 , y0 ) 6= 0. Then in some neighborhood of (x0 , y0 ), for all y there is a unique x so
that f (x, y) = z0 .
As an example, consider x2 + y 2 = 1 at the point ( 35 , 54 ). Then in some small
p
neighborhood, the function y = 1 − y 2 gives a unique x for each y.
We now prove the full Implicit Function Theorem from the Inverse Function Theorem.
Theorem 11.3 (Implicit Function Theorem)
Consider a continuously differentiable function f : U → W , where U is an open set in
the vector spaces V and W is another vector space. Let p ∈ U , and let f (p) = z. We
decompose V = V1 ⊕ V2 , and suppose that Df (p) restricted to V1 is an isomorphism
onto W .
Show that we can find neighborhoods U1 ⊆ V1 and U2 ⊆ V2 so that p ∈ U1 × U2
and the following property holds: For any b ∈ U2 we there exists a unique a ∈ U1 so
that f (a + b) = z. Moreover the function b 7→ a is continuously differentiable.
Proof. The trick is to consider the diagram
Df ⊕ id V1 ⊕ V2 W ⊕ V2
'
g (f (a + b), b)
a+b h
This function g is defined from the left to the right, but its differential (Df, id) at p is
invertible, and hence the Inverse Function Theorem applies and gives us an inverse h
with suitable restrictions.
Then given a b, we simply call h on (z, b). This gives us a + b from which we can
extract a.
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Evan Chen (Spring 2015)
11 March 10, 2015
§11.3 Differential Equations
Definition 11.4. Let U ⊆ V be open. A vector field is a function ξ : U → V .
“You should imagine a vector field as a domain, and at every point there
is a little vector growing out of it . . . What is it good for? It’s good for
differentiable equations.” – Gaitsgory
(What happens if you water a vector field?)
Definition 11.5. Let γ : (−d, d) → U be a continuous path. We say γ is a solution to
the differential equation defined by ξ if for each t ∈ (−d, d) we have
γ 0 (t) = ξ(γ(t)).
“As you go further along the real line, the grass grows taller and taller.” –
Gaitsgory
Example 11.6 (Examples of DE’s)
Let U = V = R.
(a) Consider the remarkable vector field ξ(x) = x. Then γ is a solution exactly
when
γ 0 (t) = ξ(γ(t)) = γ(t).
This has solutions γ(t) = c exp(t).
(b) If we replace ξ with ξ(x) = bx, then we are trying to solve γ 0 (t) = bγ(t), which
has solutions γ(t) = c exp(bt).
(c) If γ(x) = x2 , we are trying to solve γ 0 (t) = γ(t)2 . BLAH.
“Now we are back to theoretical math. We will not try to solve them, but we
will try to ensure that the solutions exist.
. . . And moreover they’re unique!” – Gaitsgory
In general if ξ : Rm → Rn then we can consider ξ = (ξ1 , ξ2 , . . . , ξn ), and γt = (γ1 , . . . , γn )
by consider an appropriate projection. Then the differential equation reads
γi0 (t) = ξi (γ1 (t), . . . , γn (t)).
Remark 11.7. You might see differential equations such as f 0 (t) = λ/tf (t), where t
itself appears in the right-hand side. On the problem set we will see how to convert
such time-dependent differential equations into time-independent differential equations
to make it fit into the right-hand side.
Remark 11.8. We will also be able to convert γ 00 (t) = −c/γ(t)2 into the form prescribed
above (see problem set).
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Evan Chen (Spring 2015)
11 March 10, 2015
Theorem 11.9
Let U ⊆ V , ξ : U → V and assume ξ satisfies the Lipschitz condition, meaning
0
ξ(x ) − ξ(x00 ) ≤ Λ x0 − x00 holds identically for some fixed Λ.
(a) For every x0 ∈ U there exists (−d, d) and γ : (−d, d) → U such that Dγ(t) =
ξ(γ(t)) and γ(0) = x0 . (In other words, given an initial condition γ(0) = x0
we can find a solution.)
(b) If γ1 and γ2 are two solutions and γ1 (t) = γ2 (t) for some t, then γ1 = γ2 .
Note that continuous differentiability is enough to imply the Lipschitz condition.
Example 11.10 (Counterexample if ξ is not differentiable)
2
Let U = V = R and consider ξ(x) = x 3 , with x0 = 0. Then γ(t) = 0 and
γ(t) = (t/3)3 are both solutions to the differential equation
2
γ 0 (t) = γ(t) 3 .
Proof. Let X be the metric space of continuous functions from (−d, d) to the complete
metric space B(x0 , r) (since the target space is bounded all functions are bounded). By
Problem 1 on PSet 4, we know X is a complete metric space.
We wish to use the contraction principle on X, so we’ll rig a function Φ : X → X with
the property that its fixed points are solutions to the differential equation.
Define
(
R
x0 + s∈[0,t] ξ(γ(s))
if t ≥ 0
R
Φ(γ) : t 7→
x0 − s∈[−t,0] ξ(γ(s)) if t < 0.
Ra
(The cases are just to handle the fact that we never bothered to define b for a < b. In
fact, for all future calculations in this proof we only consider the case t > 0 so one can
ignore the second line.) This function is contrived so that (Φγ)(0) = x0 and Φγ is both
continuous and differentiable. By the Fundamental Theorem of Calculus, the derivative
is exhibited by
!
Z
(Φγ)0 (t) =
ξ(γ(s)) = ξ(γ(t)).
s∈[0,t]
In particular, fixed points correspond exactly to solutions to our differential equation.
A priori this output has signature Φγ : (−d, d) → V , so we need to check that
Φγ(t) ∈ B(x0 , r). We can check that
Z
k(Φγ)(t) − x0 k = ξ(γ(s)
s∈[0,t]
Z
≤
kξ(γ(s))k
s∈[0,t]
≤ t max kξγ(s)k
s
< dA
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Evan Chen (Spring 2015)
11 March 10, 2015
where A = maxx∈B(x,r) kξ(x)k; we have A < ∞ since B(x, r) is compact. Hence by
selecting d < r/A, the above is bounded by r, so Φγ indeed maps into B(x0 , r). (Note
that at this point we have not used the fact that ξ is uniformly continuous.)
It remains to show that Φ is contracting. We wish to show that
ρ (Ψ(γ1 ), Ψ(γ2 )) ≤ λρ(γ1 , γ2 )
where ρ is the sup norm given by
ρ(f x, gx) =
sup ρX (f t, gt) .
t∈(−d,d)
Write
Z
k(Φγ1 )(t) − (Φγ2 )(t)k = (ξ(γ1 (s)) − ξ(γ2 (s)))
s∈[0,t]
Z
=
kξ(γ1 (s)) − ξ(γ2 (s))k
s∈[0,t]
≤ tΛ sup kγ1 (s) − γ2 (s)k
s∈[0,t]
< dΛ sup kγ1 (s) − γ2 (s)k
s∈[0,t]
= dΛρ(γ1 , γ2 ).
Hence once again for d sufficiently small we get dΛ ≤ 12 . Since the above holds identically
for t, this implies
1
ρ(Ψγ1 , Ψγ2 ) ≤ ρ(γ1 , γ2 )
2
as needed.
§11.4 Lawns
Let ξ be continuously differentiable. Next week we will consider a function U × R → U by
sending (x0 , t) to γ(t), where γ : (−d, d) → U is the solution to the differential equation
γ(0) = x0 (and γ 0 = ξ ◦ γ). We will show this function is nice.
From this, we will deduce that if ξ(x) 6= 0 for all x, then U can be diffeomorph’ed into
another U1 for which ξ is just a constant function. Hence,
The right way to make your lawn look nice is to apply a diffeomorphism to
the soil.
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12 March 12, 2015
§12 March 12, 2015
’Twas Housing Day. Since the last thing I wanted was to hear more about Houses, I
elected to sleep through Math 55. Notes today are again from W Mackey.
Let U ⊆ V , and ξ be a vector field. Recall that a solution to a differential equation is
a γ such that Dγ(t) = ξ(γ(t)).
Example 12.1
If ξ(x) = v for every x, then we have constant speed. So γ(t) = x + tv.
“Wouldn’t it be nice if every differential equation was like this? The good
thing is, it kind of is.” –Gaitsgory
By this, we mean that we can always diffeomorph into our constant vector field, so long
as the original vector field vanishes nowhere.
Assume there is a Λ such that ||ξ(x1 ) − ξ(x2 )|| ≤ Λ||x1 − x2 ||. Then for every x ∈ U ,
there is an interval (−d, d) and a solution γ such that γ is a solution to the differential
equation and γ(0) = x. In our
R proof last class, we actually had an algorithm: set
γ0 (t) = x, and γn+1 (t) = x + [0,t] ξ(γn (s)). Then there exist small enough d such that
each γn (−d, d) ⊆ U , and γn converges uniformly to a solution.
Proposition 12.2
For every x ∈ U , there is x ∈ U 0 ⊆ U and d such that γnmaster (x, t), defined inductively
by
γ0master (x, t) = x
γnmaster (x, t) = x +
Z
ξ(γn−1 (x, s))
are continuous functions from U 0 × (−d, d) → U that converge uniformly. Let the
limit be γ master . Then γ master is a function not just of time, but the original position.
Then γ master is continuous, and Dtime γ master (x, t) = ξ(γ master (x, t)).
Proof. This is the same as last time–d is only dependent on Λ, so by taking smaller U 0 ,
it just works.
“I could say about 60% of math is about differential equations.”
Theorem 12.3
Consider γ master (x, t) : U 0 × (−d, d) → U . If ξ is continuous differentiable n + 1 times,
then γ master will be continuously differentiable n times (possible on a smaller U 0 , d).
“The proof of this theorem I find fun... It’s kind of like the ultimate use of
the chain rule.” –Gaitsgory
Proof. To construct this, we’ll consider something else that spits it out. Set Ve :=
e T) =
e = U ⊕ Hom(V, V ), and ξ(x,
V ⊕ End(V ). Note dim(Ve ) = n + n2 . Then set U
(ξ(x), Dξ(x) ◦ T ).
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12 March 12, 2015
Matt: “Is there any intuition for why we’re doing this?”
Gaitsgory: “That’s a great question. No.” (This is because we get seemingly
unnatural choices since we’re doing a special case of the theorem. It makes
far more sense we we do the general theorem on manifolds.)
Then there is
e 0 × (−d, d) → Ve
γ master : U
e We’re interested in (x, idV ). Then
is a solution to ξ.
γ
emaster : U 0 × (−d, d) → V × End(V )
is given by (0 γ master (x, t), δ(x, t)). We claim that 0 γ master = γ master . We know γ
emaster
satisfies
• Dtime (0 γ master (x, t)) = ξ(0 γ master (x, t))
• Dtime δ(x, t) = Dξ(0 γ master (x, t)) ◦ δ(x, t)
Then
e γ master (x, t)),
Dtime γ
emaster (x, t) = ξ(e
and
γ
emaster (x, t) = (0 γ master (x, t), δ(x, t)).
Then just follow through on the components. Note that 0 γ(x, 0) = x, so we have the
same initial condition, hence, 0 γ = γ. From here, we’ll deduce the theorem from the next
theorem.
Theorem 12.4
We have δ(x, t) = Dspace (γ(x, t)).
Corollary 12.5
Theorem 12.4 implies
Dtime Dspace γ(x, t) = Dξ(γ(x, t)) ◦ Dspace (γ(x, t)).
We now claim that Theorem 12.4 implies Theorem 12.3: We’ll proceed by induction
on n. For the base case, we want to show that the partials are continuously differentiable.
Note Dtime γ(x, t) = ξ(γ(x, t)) is continuous, immediately. Then Dspace γ(x, t) = δ(x, t) is
continuous since it’s a component of γ
e.
Now suppose we know the theorem for n, and ξ is differentiable n + 2 times. We want
γ to be continuously differentiable n + 1 times. Then we have Dtime γ differentiable n
times and Dspace γ differentiable n times. Then just take the derivatives and apply the
hypothesis, it works. It just remains to prove Theorem 12.4.
Proof. Set γ0 (x, t) = x, δ0 (x, t) = id, and let
Z
γn+1 = x +
t
Z
ξ(γn (x, s)) δn+1 = id +
0
0
53
t
Dξ(γn (x, s) ◦ δn (x, s).
Evan Chen (Spring 2015)
12 March 12, 2015
Then we claim that Dspace γn (x, t) = δn (x, t). We know that
uniformly
γn −−−−−−→ γ
and
uniformly
δn −−−−−−→ δ
so by our theorem on the derivative of a limit, Dspace γ = δ, so long as it’s true on each
component. Now by induction–it’s true on the base case. Suppose it’s true for n. Then
Z
Z
Dspace γn+1 (x, t) = Dspace (x + ξ(γn (x, s)) = id + Dspace ξ(γn (x, s)),
which by midterm problem 2, is
Z
id +
Dspace ξ(γn (x, s)).
Then we want
Dspace (ξ(γn (x, s))) = Dξ(γn (x, s)) ◦ δn (x, s).
This is just the chain rule and the inductive hypothesis, however.
Theorem 12.6
Let ξ be a vector field such that ξ(x) 6= 0 at x. There there is U 3 x with
f
e ⊆ V 0 × R such that f is a diffeomorphism and f transforms ξ to the vector
U−
→U
field (0, 1).
Definition 12.7. Let
U
∩
V
f
e
U
∩
Ve
with f a diffeomorphism (both it and its inverse are differentiable), and ξ be a vector
e ) = Df (x)(ξ(x)). Then we say that f transforms ξ to
field on V . Then suppose ξ(V
e
the vector field ξ. This is a sensible map, in the sense that it maps solutions to the
differential equation to other solutions of the differential equation.
Proof. Assume ξ is continuously differentiable, and γ master : U 0 × (−d, d) → U . Let
V 0 ( V be such that
V 0 ⊕ Span(ξ(x)) = V.
Then define g : V 0 × (−d, d) → U by
g(v 0 , t) = γ master (x + v 0 , t).
so Dg(0, 0) is an invertible map from V 0 × R → V . Then Dtime g(0, 0) = ξ(x), and
Dspace g(0, 0) = δ(x) = id. Hence the Inverse Function Theorem applies. Then we
want that g transforms the constant vector field (0, 1) to ξ. That is, we want to show
Df (x0 , t)(0, 1) = ξ(g(v 0 , t)). We have
g(v 0 , t + s) − g(v, t)
= Dtime γ master (v 0 , t)
s→0
s
= ξ(γ master (v 0 , t))
lim
as desired.
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13 March 24, 2015
§13 March 24, 2015
§13.1 Midterm Solutions
3. Let γ : [0, 1] → V be a continuous function such that its derivative γ 0 : (0, 1) → V
extends to a continuous function on [0, 1]. Fix a norm on V , and assume the following
additional property:
||v1 + v2 || = ||v1 || + ||v2 |, ⇒ v1 and v2 are proportional.
Show that the following conditions are equivalent:
• `(γ) = ||γ(1) − γ(0)||.
• There exists v ∈ V such that γ 0 (t) = a(t) · v for all t ∈ (0, 1) with a(t) ∈ R≥0 .
• There exists v ∈ V such that γ(t) = γ(0) + b(t) · v for all t ∈ [0, 1] with b(t)
satisfying t1 ≤ t2 ⇒ b(t1 ) ≤ b(t2 ).
The only hard part is showing that the first implies the third. If not, we have
kγ(1) − γ(0)k < kγ(1) − γ(t)k + kγ(t) − γ(0)k ≤ `(γ : 0 → t) + `(γ : t → 1) = `(γ).
This is a contradiction.
2. Let f be a continuous function [a1 , b1 ]×(a2 , b2 ) → R, such that for any t1 ∈ [a1 , b1 ],
the function f (t1 , −) : (a2 , b2 ) → R is differentiable, and the resulting function
∂2 f (t1 , t2 ) : [a1 , b1 ] × (a2 , b2 ) → R
is continuous. Consider the function
Z
F : (a2 , b2 ) → R,
F (t2 ) :=
f (t1 , t2 ).
t1 ∈[a1 ,b1 ]
Show that F is differentiable and
0
Z
F (t2 ) =
∂2 f (t1 , t2 ).
t1 ∈[a1 ,b1 ]
The point is to show that
R
0 = lim
h→0
s∈[a,b] f (s, t
+ h) −
R
s∈[a,b] f (s, t)
h
−
!
Z
∂t f (s, t)
s∈[a,b]
Fix ε > 0 and let δ be such that
ρ (t01 , t02 ), (t1 , t2 ) < δ =⇒ ∂2 f (t01 , t02 ) − ∂2 f (t1 , t2 ) <
We claim that
f (t1 , t02 ) − f (t1 , t2 )
ε
− ∂2 f (t1 , t2 )
0
≤ b1 − a1 .
t2 − t2
ε
.
b−a
By the Mean Value Theorem there is a |tildet2 such that the fraction above equals
∂2 f (t1 , t˜2 ) − ∂2 f (t1 , t2 ).
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13 March 24, 2015
1. (Arzela-Ascoli) Let X and Y be compact metric spaces, and let {fn } be a sequence
of elements in Functcont (X, Y ). Assume that the sequence satisfies the following
additional hypothesis:
For every there exists δ such that for ρ(x1 , x2 ) < δ we have ρ(fn (x1 ), fn (x2 )) <
for every n.
Show that under this hypothesis, the sequence {fn } does have a convergent subsequence.
Let X be compact, and let x1 , x2 , . . . be dense in X (possible since X is compact).
Lemma 13.1
The sequence (fn ) contains a subsequence gn such that gn (xi ) converges.
Proof. Uses a diagonal argument.
Now we claim that the gn are Cauchy. For all ε > 0 we seek N such that for all X, we
have
kgn (x) − gm (x)k < ε.
Let δ be such that ρ(x0 , x00 ) < δ implies ρ(fn x0 , fn x00 ) <
continuous assumption). In particular,
1
3ε
(here we use the equi
1
ρ(gn (x0 ), fn (x00 )) < ε ∀n.
3
There exist finitely many balls of radius 12 δ that cover X, since X is compact. Thus there
exists for each j a point from our dense set xij . Hence for all x there exists j such that
ρ(x, xij ) < δ.
The rest is clear.
§13.2 PSet Review
7. Let ξ be a vector field on a domain U ⊂ V . Let U 0 ⊂ U and (−d, d) ⊂ R be such
that there exists a function
γ master : U 0 × (−d, d) → U
that satisfies:
• Dtime γ master (x, t) = ξ(γ(x, t));
• γ master (x, 0) = x.
In this case, we shall say that the solution to the differential equation defined by ξ is
defined on U 0 × (−d, d). For −d < t < d denote
φt (x) = γ master (x, t),
viewed as a function U 0 → U .
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Evan Chen (Spring 2015)
13 March 24, 2015
(a) Let 0 < d0 ≤ d and U 00 ⊂ U 0 be such that γ master (x, t) ∈ U 0 for x ∈ U 00 and
−d0 < t < d0 . Show that the solution
γ master : U 00 × (−d − d0 , d + d0 ) → U
is defined and the corresponding maps φt satisfy
φt+t0 (x) = φt (φt0 (x)) for − d0 < t0 < d0 and − d < t < d.
Because of the above property, we call the maps φt the flow associated to ξ.
Fix x. To show that it’s defined for t < d + d0 , note that γ master (x, t − d) ∈ U 0 , and
so we may apply the flow again (shifted by d) to define γ master (x, t) ∈ U . Similarly for
t > −(d + d0 ).
(b, (mandatory, but still bonus 1pt)) Assume that ξ is differentiable (n + 1) times
with n ≥ 1. Show that γ master is differentiable n times and
Dtime Dspace γ master (x, t) = Dξ(γ master (x, t)) ◦ Dspace γ master (x, t).
NB: in class we proved the above assertion for some U 0 and d small enough. So, in
this problem you’re being asked to go from small (U 0 , d) to any (U 0 , d) on which the
master solution is defined.
Abbreviate γ = γ master . We have a function
Dspace γ : U 0 × (−d, d) → Hom(V, V ).
We want to check
Dtime Dspace γ(x, t) = Dξ(γ(x, t)) ◦ Dspace γ(x, t).
Observe that the partials commute, so we can rewrite the LHS as
Dspace Dtime γ(x, t) = Dspace ξ(γ(x, t)) = (Dspace ξγ(x, t)) ◦ Dspace γ(x, t)
the last step following from the chain rule.
(c, (mandatory, but still bonus 2pts)) Show that the maps φt : U 0 → U have an
invertible differential (see Problem 6) for every −d < t < d at every x ∈ U 0 .
It’s true when t = 0, and hence for small neighborhoods (say by looking at matrices).
Then use compactness.
8. Let ξ1 and ξ2 be vector fields on a domain U ⊂ V , both continuously differentiable
twice. Let U 0 ⊂ U and 0 < d be such that master solutions both γξmaster
and γξmaster
2
1
0
for both ξ1 and ξ2 are defined on U × (−d, d).
Let U 00 ⊂ U 0 let 0 < d0 ≤ d be such that for i = 1, 2 we have
γξmaster
(x, t) ∈ U 0 for x ∈ U 00 and − d0 < t < d0 .
i
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Evan Chen (Spring 2015)
13 March 24, 2015
In particular, the expressions
(x, t2 ), t1 )
(γξmaster
(x, t1 ), t2 ) and γξmaster
(γξmaster
γξmaster
2
1
1
2
are defined for x ∈ U 00 and −d0 < t1 , t2 < d0 .
(a, (mandatory, but still bonus 2pts)) Assume that [ξ1 , ξ2 ] = 0. Show that
(x, t2 ), t1 )
(γξmaster
(x, t1 ), t2 ) = γξmaster
(γξmaster
γξmaster
2
1
1
2
for every x ∈ U 00 and −d00 < t1 , t2 < d00 for some 0 < d00 ≤ d0 .
NB: in what follows we shall say that two vector fields ξ1 and ξ2 commute if
[ξ1 , ξ2 ] = 0. Thus, the above problem says that if vector fields commute, then the
flows that they define commute as well.
Hint: reduce to the case when one of the two fields doesn’t vanish at x and use
the straightening theorem.
If both fields vanish at x, then both sides of the equation are equal to x (nothing moves).
Otherwise, assume ξ1 doesn’t vanish, take U 00 a neighborhood at x non-vanishing, and
transform to a situation where ξ1 is a constant vector field with, say, value e1 (in Ṽ ).
Then this amounts to showing that
γξmaster
(x + e1 t1 , t2 ) = γξmaster
(x, t2 ) + e1 t1 .
2
2
According to Problem 4(e), [ξ1 , ξ2 ] = 0 implies that ξ2 is independent of the first
coordinate, so this follows.
9. Let ξ be a vector field on a domain U ⊂ V , continuously differentiable twice. Let
U 0 ⊂ U and (−d, d) ⊂ R be such that
γ master : U 0 × (−d, d) → U
is defined.
For another continuously differentiable vector field η on U , consider the (timedependent) vector field ηt on U 0 , equal to φ∗t (η) (see Problem 6). I.e., we regard ηt
as a function
Ux × (−d, d) → V.
The φt is reversible by 7(c). So we want to think of a vector field
(φt )∗ (η) : U 0 × (−d, d) → V.
(b, (mandatory, but still bonus 2pts)) Consider the vector field Dtime ηt (x, 0) on U 0 .
Show that
Dtime ηt (x, 0) = [ξ, η](x).
Hint: use your knowledge of Dtime Dspace γ master (x, t).
Chain Rule. The left-hand side is
Dtime (Dφt )−1 (ηφt x) .
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Evan Chen (Spring 2015)
13 March 24, 2015
Here Dφt is the matrix and ηφt x is a vector. Using the general product rule at t = 0, we
get
Dtime (Dφt )−1 (ηφ0 x) + (Dφt )−1 Dtime (ηφt x) = Dtime (Dφt )−1 (ηx) + Dtime (ηφt x).
Recall that we want the answer to be Dξ η(x) − Dη ξ(x).
First, the Chain Rule gives
Dtime (ηφt x). = Dη(x) ◦ ( Dtime φt (x)|t=0 ) = Dη(x) ◦ ξ(x)
the last step following from the differential equation. By definition, this equals Dξ η(x).
Now, we want to show
Dtime (Dφt )−1 = −Dξ(x).
t=0
Once this is done we can substitute in directly. We use the trick that
Dtime A(t)−1 = −Dtime (A(t)) .
This follows from the fact that
0 = Dtime A(t) · A(t)−1
and applying the Leibniz rule. So, it only remains to compute that
Dtime (Dφt )|t=0 = Dtime Dspace γ|t=0
= Dξ(γ(x, t)) · Dspace γ(x, t)|t=0
= Dξ(γ master (x, 0)) · Dspace idV
= Dξ(x).
5. Let ξ1 , ξ2 , ξ3 be vector fields, each continuously differentiable twice. Prove the
Jacobi identity
[[ξ1 , ξ2 ], ξ3 ] + [[ξ3 , ξ1 ], ξ2 ] + [[ξ2 , ξ3 ], ξ1 ] = 0
(preferably, without long formulas).
Appeal to part (f) previously. We have
[[ξ1 , ξ2 , ] , ξ3 ] = D[ξ1 ,x2 ] ξ3 − Dξ3 [ξ1 , ξ2 ]
= [Dξ1 , Dξ2 ] (ξ3 ) − Dξ3 (Dξ1 ξ2 − Dξ2 ξ1 )
= Dξ1 Dξ2 ξ3 − Dξ2 Dξ1 ξ3 − Dξ3 Dξ1 ξ2 + Dξ3 Dξ2 ξ1 .
Cyclically summing yields the conclusion.
§13.3 Review of Exterior Products
Recall the definition of Λk (W ) in the “quot” form and define
M
Λ• (W ) =
Λk (W ).
k
Moreover, given w1 , . . . , wn ∈ W we let
w1 ∧ · · · ∧ wn
denote the image of w1 ⊗ · · · ⊗ wn .
In addition, given α ∈ Λn1 (W ) and β ∈ Λn2 (W ), we may put
α ∧ β = (−1)n1 n2 β ∧ α ∈ Λn1 +n2 (W ).
Finally, recall that if e1 , . . . , em is a basis of W then a basis of Λn (W ) is
ei1 ∧ · · · ∧ ein i1 < · · · < in .
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Evan Chen (Spring 2015)
13 March 24, 2015
§13.4 Differential Forms
Let U ⊆ V be a domain. Let Ωn (U ) denote the set
C ∞ (U, Λn (V ∨ )).
L
Define Ω• (U ) = n Ωn (U ).
Now suppose we have a map
φ : U1 → U2
of domains. We define a map
φ∗ : Ωn (U2 ) → Ωn (U1 )
by sending each α ∈ Ωn (U2 ) to
def
(φ∗ (α)) (x) = Λn ((Dφ)(x))∨ (α(φ(x))) = Λn [(α ◦ φ)(x) ◦ (Dφ)(x)]
Here, given T : W1 → W2 we have Λn (T ) = T ∧ · · · ∧ T .
In particular, if α is a 0-form, then a simpler definition is given by
φ∗ (α) = α ◦ φ.
Also,
Theorem 13.2 (de Rham Derivative)
d
There exists a unique map Ωn −
→ Ωn+1 (U ) (for all n) with the following properties.
• For f ∈ Ω0 (U ), df = Df .
• d(α ∧ β) = (dα) ∧ β + (−1)deg α α ∧ (dβ).
• d ◦ d(α) = 0.
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Evan Chen (Spring 2015)
14 March 26, 2015
§14 March 26, 2015
I managed to oversleep class today, so again these are from W.
“Today I’m going to do something pretty reckless...” –Gaitsgory
§14.1 More on Differential Forms
Let U ⊆ V ; recall we set Ω0 (U ) = C ∞ (U ), Ω1 (U ) = C ∞ (U, V ∨ ), and in general, Ωk (U ) =
C ∞ (U, Λk V ∨ ). Given two wedges of functions, we can define (α ∧ β)(x) = α(x) ∧ β(x),
and (f · α)(x) = f (x) · α(x).
Problem 1 on this week’s PSet is saying that there exists a d : Ωk (U ) → Ωk+1 (U ) that
satisfies
• d(f ) = Df
• d(α ∧ β) = dα ∧ β + (−1)deg α α ∧ dβ.
• d(d(α)) = 0.
So for instance, on 1-forms: take f, g ∈ Ω1 (U ). Then d(f dg) = df ∧ dg + f ∧ d(d(g)) =
df ∧ dg.
Lemma 14.1
Let α be a 1-form.
(a) The form α is the sum of 1-forms of the shape f dg. If V = Rn , α is uniquely
of the shape f dg.
P
(b) If V = Rn , α is uniquely of the shape
fi dxi .
P
∨
∨
fi ei .
Proof. (b) clearly implies (a). Let e∨
1 , ..., en be a dual basis for V , then let α =
∨ 3 ψ, consider the 1-form
Note the 1-form dxi acts by dxi (x) = e∨
.
More
generally,
if
V
i
{αψ , fψ (x) = ψ}. The we claim that for the function fψ , dfψ = αψ . Note that dfψ = Dfψ
since it’s a 0-form. Thus Dfψ (x) : V → R. But then for any linear map T : V → V 0 , we
have DT (x) = T , hence Dfψ (x) = ψ.
Given
V1
V2
U1
ϕ
U2
we have a map ϕ∗ : C ∞ (U2 ) → C ∞ (U1 ). Moreover, for a k-form α, (ϕ∗ (α))(x) ∈ Λk (V1∨ ).
This is defined sensibly: we have for x ∈ U1 that
Dϕ(x)
V1 −−−−→ V2
(Dϕ(x))∨
V1∨ ←−−−−−− V2∨
Λk (Dϕ(x)∨ )
Λk (V1∨ ) ←−−−−−−− Λk (V2∨ )
and of course, α(ϕ(x)) ∈ Λk (V2∨ ).
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§14.2 Topological Manifolds
“Now, prepare for the worst.” –Gaitsgory
Definition 14.2. A topological manifold of dimension n is a Hausdorff topological
space such that for every x ∈ X, there is an open set x ∈ Ux ⊆ X such that there is a
homeomorphism from Ux to U ⊆ Rn .
Example 14.3
Any U ⊆ Rn .
Example 14.4
Take S n ⊆ Rn+1 , with a homeomorphism given by projections from the north pole.
(To be argued and clarified later.)
We’ll now define C ∞ manifolds. Apparently, we aren’t risking losing any generality by
ignoring the C k case.
Definition 14.5. A C ∞ manifold is a topological space equipped with a collection
Uα
ϕ
U
X
V
with ϕ a homeomorphism such that
1. For every x ∈ X, there is a chart that contains it.
ϕα
ϕβ
2. Given Vα ⊇ Uα ' U 0 ,→ X ←- U 00 ' Uβ ⊆ Vβ , consider U 0 ∩ U 00 ⊆ X. Then
ϕα
ϕβ
−1
0
00
0
00
−1
0
00
0
00 ∼
ϕ−1
α (U ∩ U ) ⊆ Uα and ϕβ (U ∩ U ) ⊆ Uβ , so ϕα (U ∩ U ) ∼ U ∩ U
0
00
∞
ϕ−1
β (U ∩ U ). We require that this homeomorphism is C . Terminologically, we
∞
say that charts intersect in C , or are smoothly compatible. It is worth noting
that ϕ−1
α ◦ ϕβ : Uβ → Uα is a map from domains of vector spaces, so it makes sense
to ask whether the homeomorphism is C ∞ .
3. This is just a completeness condition: suppose we have Uγ ⊆ Vγ equipped with a
homeomorphism γ between it and U ⊆ X, and this is smoothly compatible with
all of the charts, then Uγ is itself a chart.
Conditions (1) and (2) define an atlas; it is a maximal atlas if (3) holds.
“If you pick up a piece of paper, and it looks like a chart and fits in your
folder, you can put it in your folder since it is.”
Lemma 14.6
Let X be a Hausdorff topological space, equipped with a set of charts satisfying (1)
and (2). Then X admits a unique structure of C ∞ manifold such that (Uα , ϕα ) are
charts.
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Evan Chen (Spring 2015)
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Proof. Take any
Uγ
ϕγ
U
X
Vγ
We want to include it if and only if it’s compatible with current charts. Then we
declare (Uγ , ϕγ ) to be a chart if it’s C ∞ against our current charts. Then completing the
construction is more or less tautological, and therefore on the PSet.
§14.3 Smoothness of Manifolds
Now, back to our example of S n , we can just specify an atlas. Take planes both above and
below the sphere, and lines from both the north and the south pole. Then we have charts
for S n −{south pole} (“No polar bears. Sad.” –Gaitsgory) and S n −{north pole}, each of
which are homeomorphic to Rn . Then we get Rn − {0} ' S n − {N.P & S.P.} ' Rn − {0}
with respect to both projections.
“No polar bears or penguins, now.”
Then we want to claim that these homeomorphisms are C ∞ . You can just do it, but
it’s ugly. There’s a slicker way.
Let X be a C ∞ manifold.
Definition 14.7. A function f on X is said to be C ∞ if it’s C ∞ on each chart: f |U ◦ ϕα :
Uα → R.
Definition 14.8. A k-form on X, α ∈ Ωk (X), is a datum of a k-form on every chart
ϕ0
ϕ0
e 0 ⊆ V 0 and U 00 '
e 00 ⊆ V 00 , with α0 a k-form on U
e 0 and
such that for U 0 , U 00 ⊆ X, U 0 ' U
U
00
00
0
00
e
α a k-form on U , ϕ and ϕ are compatible in the sense that
ϕ0−1 (U 0 ∩ U 00 )
ϕ0 (ϕ00 )−1
e0
U
(ϕ00 )−1 (U 0 ∩ U 00 )
e 00
U
Then we want to require
(ϕ0 ϕ00−1 )∗ (α0 |ϕ−1 (U 0 ∩U 00 ) ) ' α00 |ϕ00−1 (U 0 ∩U 00 ) .
A vector field on a manifold is is the same thing: We have ξ(x) ∈ V , α(x) ∈ Λk (V ∨ ).
But our definition doesn’t really give this–it gives different things on each chart. If X is
a differentiable manifold, x ∈ X, we’ll introduce Tx X, the tangent space at x, and Tx∨ X,
the cotangent space.
§14.4 Cotangent space
Take x ∈ U ⊆ X, f a C ∞ function. Df (x) will live in the cotangent space, once we
define it.
Definition 14.9. We set Df (x) = 0 if it’s 0 on some/every chart.
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Evan Chen (Spring 2015)
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Definition 14.10. Fix x ∈ X. A germ is a pair (U, f ), where f : U → R is smooth. We
say two germs (U1 , f1 ) and (U2 , f2 ) are equal if f1 and f2 agree on some neighborhood of
U1 ∩ U2 : that is, a germ only cares about whatf does on arbitrarily small neighborhoods
of x.
One can endow this with a vector space structure in the obvious way, and observe that
it makes sense to talk about f x and (Df )x for a U ⊆ V .
Definition 14.11. Define Tx∨ X to be the set of germs vanish at x modulo the set of
germs whose differential vanish at x. This is the cotangent space of x at V . We want
that this is V ∨ on each chart.
Lemma 14.12
Let U be contained in a chart. If U ⊆ V , then Tx∨ X ' V ∨ .
Proof. Define this map by f 7→ Df (x). This is well defined and exactly kills the quotient.
Thus injectivity is by definition. Then to verify that we have every element, for any
functional f ∈ V ∨ , define a function g(v) = f (v) − f (x). Then this is in TX∨ , and
Dg = f ∈ V ∨ since that’s true for all linear functions.
We can then define the tangent space Tx X = (Tx∨ X)∨ . More concretely, consider
γ
→ X so that γ(0) = x. Then we’ll mod out by an equivalence relation ∼
paths (−, ) −
when two paths are tangent to each other at x.
Definition 14.13. Let X and Y be C ∞ manifolds. A C ∞ map f : X → Y is a
continuous map such that for every x, we can consider the function f (x) ∈ Uy ⊆ Y ,
ex = f −1 (Uy ), and then Ux to be U
ex
with Uy homeomorphic to Uy0 . Then we can set U
0
intersected with some chart around x, possibly completely containing Ux , so that we have
Ux
f
∼
Ux0
Uy
∼
f0
Uy0
with Ux0 ⊆ Vx0 and Uy0 ⊆ Vy0 . Then we ask for f 0 to be C ∞ .
Now, we say γ1 ∼ γ2 if “Dγ1 (0) = Dγ2 (0).” We don’t know what either side of the
equation is because we don’t even know where they lie. However, we know what it means
for them to be equal, since that just means they’re equal on some/all charts, so we’re OK.
Now, finally, we can claim Paths / ∼' Tx X. “This is easy to prove, but it will overflow
your minds if I do it. . . Too much for now.” –Gaitsgory
(In an email about the PSet) “The problems on manifolds are all more or less
tautological, once you figure out what they are talking about.” – Gaitsgory
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§15 March 31, 2015
Here is a joke.
Old McDonald had a form
ei ∧ ei = 0.
Today Aaron Landesman is teaching integration. He says: “today we’ll make a mess
and just and define things without checking that they are well-defined. On Thursday
Ashwin will get to clean up the mess and check things are well-defined”.
§15.1 Boxes
Let S = [a1 , b1 ] × · · · × [an , bn ]. Let f : S → R be uniformly continuous1 . Take partitions
pi with increasingly small meshes.
For each partition p into hypercubes {Sα } we can take sample points in the boxes xα ,
and consider
X
Σp =
f (xα ) vol(Sα )
α
and define
Z
f=
S
lim
mesh p→0
Σp .
Proposition 15.1
This is well-defined, like the Riemann integral case.
Proof. Left for Ashwin on Thursday.
“Left as an exercise to the Ashwin.” – James Tao
§15.2 Supports
Let f : U → R where U ⊆ Rn .
Definition 15.2. We say the support of f , denoted supp(f ), is the closure in Rn of
the set {x | f (x) 6= 0}.
Note that the support may lie outside of U .
Definition 15.3. Suppose f : U → R is uniformly continuous and the support of f is a
subset of U ⊆ Rn , we define f : supp(U ) → R by
(
f (x) x ∈ U
f (x) =
0
x∈
/ U.
1
Or just continuous, since S is compact. But in general today our functions are going to be uniformly
continuous.
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Evan Chen (Spring 2015)
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Lemma 15.4
In the above notation, f is continuous.
Proof. It’s trivial if x ∈ U ,and for x ∈
/ U use the fact that U \ supp(f ) is closed.
Definition 15.5. Assume U is bounded, and supp(f ) ⊆ U . Let S be a box containing
U . Then we define
Z
Z
f=
f.
U
S
§15.3 Partitions of Unity
Definition 15.6. Let A ⊆ Rn , and let {Uα } be an open cover of A such that Uα is
bounded. Then a partition of unity subordinate to {Uα } is defined as follows. We pick
an open set U ⊇ A an open set, and a set of smooth functions U → [0, 1], denoted Φ. It
must satisfy the following properties.
(a) For all x ∈ A, there is a neighborhood Vx 3 x, such that only finitely many φ ∈ Φ
are nonzero when restricted to Vx .
P
(b)
φ∈Φ φ(x) = 1 for each x ∈ A (this sum is finite by the preceding condition).
(c) For any φ ∈ Φ there exists an α such that supp(φ) ⊆ Uα .
In a moment we’ll show these exist. Once we do we can defie the following.
Definition 15.7. Let A ⊆ Rn , and f : A → R be uniformly continuous. Then we define
Z
XZ
f=
φ·f
A
φ∈Φ A
where {Uα } is an open cover of A and Φ is a partition of unity subordinate to {Uα }.
Theorem 15.8
This definition does not depend on the choice of {Uα } and Φ. It also does not depend
on the order of the infinite sum.
§15.4 Existence of Partitions of Unity
Theorem 15.9
Let A ⊆ Rn and let Uα be an open cover of A such that each Uα is bounded. Then
there exists a partition of unity subordinate to it.
The remainder of the lecture is devoted to the proof of this.
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§15.5 Case 1: A is Compact
Let’s first solve the case if A is a compact set.
Lemma 15.10
Given B(p, r) a ball then we may select s < r and f : Rn → [0, 1] a smooth function
such that
(
1 p ∈ B(x, s)
f (x) =
0 p∈
/ B(x, r).
Proof. Choose s sufficiently small so that there are cubes R and S centered at x such
that
B(x, r) ⊃ R ⊃ S ⊃ B(x, s).
(Actually I think if you equip Rn with the max metric this is clear.) Now for n = 1 we
can use a bump function toQget the desired result.
Q For n > 1, just take a product of these
smooth functions: if R = i [ai , bi ] and S = i [ci , di ], define gi : R → R to work on the
ith dimension, and let
Y
f (x1 , . . . , xn ) =
gi (xi ).
i
We’re given Uα a cover of A. Take a ball Bx ⊆ Uα for each x ∈ X, and let Bx0 be a
ball within each Bx as described by the lemma: that means we have a smooth function
fx for each X such that
• fx is one on the set Bx0 , and
• fx is zero outside of Bx .
S
Now Bx0 = A, so we can take a finite subcover of size n. Denote the corresponding
balls and functions by Bi0 and fi .
Then, for each i we set
fi
φi =
f1 + · · · + fn
as a function A → R.
Proposition 15.11
The φi are a partition of unity on
S
i Ui
⊃ A which is subordinate to Uα .
P
Proof. First, observe that i fi > 0 for every x ∈ X, since x ∈ Bi0 for some i and thus
fi (x) = 1 > 0. Hence the denominator never vanishes, and so φi is indeed a function to
[0, 1].
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Evan Chen (Spring 2015)
15 March 31, 2015
fi
Moreover, φi = f1 +···+f
< 1 for every x.
n
Finally, we see that the support of each φi lives within Ui by construction.
§15.6 Case 2: A is a certain union
We have thus solved the problem when A is compact. Now, we will prove the result in
the case
[
A=
Ai
i≥1
with Ai compact, and so that the interior of Ai+1 contains Ai for which each Ai . This is
the hardest case, so one might want to skip to the next subsection first.
Define
Bi = int(Ai+1 ) \ Ai−2
and
Ci = Ai \ int(Ai−1 ).
Hence Ci are compact and Bi ⊃ Ci . Take for each i an open cover
i
Uα α = {Uα ∩ Bi }α
an open cover of Bi . (At this point we can forget about the Bi now; there’s just an open
set for the Ci to live in).
Then there exists a partition of unity Φi of Ci which is subordinate to Uαi . Now we set
[
Ψ=
Φi
where all funcitons are now extended by zero to all of Rn . Moreover, we define
X
σ(x) =
ψ(x).
ψ∈Ψ
Lemma 15.12
The above σ is well-defined in the sense that only finitely many ψ(x) terms have
nonzero contribution. Moreover, σ(x) is never zero.
Proof. We wish to check there are only finitely many ψ ∈ Ψ such that ψ(x) 6= 0. Fix x,
x ∈ Ai \ Ai−1 . Observe that if ψ ∈ Φk for some k > i + 2, we have ψ(x) = 0. So we only
need to consider k < i + 2, each of which there are finitely many, so we’re done.
The fact that σ(x) 6= 0 for any given x ∈ A, we know that x ∈ Ai for some i, at which
point the Ψi contributes a nonzero term.
Now, we define
φi (x) =
In this way,
as usual.
P
φ∈Φ φ
Ψi (x)
σ(x)
= 1. Moreover, the support condition follows from the previous case
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Evan Chen (Spring 2015)
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Example 15.13
Work in R. If A1 = [0, 1], A2 = [0, 2], A3 = [0, 3], and A4 = [0, 4], then
C3 = [2, 3] ⊆ B3 = (1, 4).
Φ3
B3
C3
A1
A2
A3
A4
NOT TO SCALE sheesh (The blue guys are supposed to be a unity partition and they’re
not.)
The “onion ring” guarantees that there are finitely many nonzero φ’s at a given point,
and the σ is just normalizing.
§15.7 Case 3: A open
Definition 15.14. The boundary of a set A is defined as the difference of the closure
and the interior:
A \ int A.
We show we can reduce the previous case now.
Define
1
Ai = x | d(x, ∂A) ≥ , |x| ≤ i .
i
Example 15.15
Take R2 and let U = {(x, y) | |y| < 2.2} be a horizontal strip. Then this might look
the following.
0
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§15.8 Case 4: General A
Given a general A, consider an open
S cover Uα of A. We know there exists a partition of
unity subordinate to Uα of U = α Uα . Then Φ is a partition of unity for A. (Check
this.)
This completes the proof.
“Now we know that partitions of unity exists. You basically don’t need to
know this proof.”
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§16 April 2, 2015
Ashvin gets to clean up the mess now.
§16.1 Integration on Bounded Open Sets
First, consider a rectangle S = [a1 , b1 ] × · · · × [an , bn ], and let f : S → R be continuous
(and hence uniformly so).
We want to show the definition we gave last time is well-defined. Recall that for each
partition p into hypercubes {Sα } we can take sample points in the boxes xα , and consider
X
Σp =
f (xα ) vol(Sα )
α
and define
Z
f=
S
lim
mesh p→0
Σp .
We need to show that
• this limit exists,
• it’s independent of the choice of sample points, and
• it’s independent of the choice of partitions.
Note that f is uniformly continuous, as S is compact.
Proof that the limit exists. We show that it’s Cauchy.
Pick ε > 0. Since f is uniformly continuous, there is a δ > 0 such that whenever
ε
kx − x0 k < δ we have |f (x) − f (x0 )| < vol(S)
. Note that there exists N such that for all
n > N and α ∈ Pn , there exists a ball B of diameter δ which completely contains the
square containing α in pn .
We claim this N works. Let m, n > N . Let p0 be the partition obtained by superposing
pm and pn , and let
X
Σ0 =
f (xα ) vol(Sα ).
α∈p0
We claim that Σ0 differs from Σm and Σn by less than ε, which will give |Σm − Σn | < 2ε
which is certainly sufficient.
This is some direct computation:
X
0
X
m
m
Σ − Σm = f (xα ) vol(Sα ) −
f (xβ ) vol(Sβ )
α∈p0
β∈pm



X
X
m
m

= f (xα ) vol(Sα ) − f (xβ ) vol(SB )
β∈pm
α∈β



X
X
X
m




=
f (xα ) vol(Sα ) − f (xβ )
vol(Sα ) β∈pm
α∈β
α∈β
X X
m
=
f (xα ) − f (xβ ) vol(Sα ) β∈pm α∈β
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Evan Chen (Spring 2015)
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Applying the triangle inequality,
X X
f (xα ) − f (xm ) vol(Sα )
≤
β
β∈pm α∈β
<
X X
β∈pm α∈β
ε
vol(Sα )
vol(S)
= ε.
.
Proof of independence of sample points. Fix a choice of partitions. Suppose we pick
sample points xα and x0α to obtain Σn and Σn0 . We show that |Σn − Σ0n | → 0. Fix ε > 0
and select δ > 0 so that points within δ have f -output within ε (uniform continuity of
f ). Let the diameter of the boxes be less than δ for n ≥ N . Then for such n we obtain
X
X
Σn − Σ0n ≤
f (xα ) − f (xα )0 vol(Sα ) ≤
ε · vol(Sα ) = ε vol(S).
α
α
Proof of Independence of Partition Choice. There’s actually a nice trick to show independence of the choice of partitions. Suppose that {pn } and {qn } are two such partitions.
Interlace the sequences
p1 , q1 , p2 , q2 , . . .
Then the limit exists to some limit L, and hence the subsequences pn and qn both
converge to L.
R
Hence it makes sense to define S f = limn→∞ Σn .
§16.2 This Integral is Continuous
For now on, for simplicity, let
S = [a1 , b1 ] × [a2 , b2 ] ⊆ R2 .
In particular we’re just working in two dimensions now, so that my LATEX-ing is easier.
Let f : S → R be (uniformly) continuous. Define g : [a1 , b1 ] → R by
Z
g(x) =
f (x, y).
y∈[a2 ,b2 ]
Analogously, define [a2 , b2 ] → R by
Z
h(y) =
f (x, y).
x∈[a1 ,b1 ]
Lemma 16.1
The functions g and h are continuous.
Proof. We’ll just do g since h is analogous. . . Pick x ∈ [a1 , b1 ]. Consider an interval
I = [x − r, x + r] ⊆ [a1 , b1 ] (though for the n-dimensional case I is a closed ball of radius
r). Then f is uniformly continuous on I × [a2 , b2 ].
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Evan Chen (Spring 2015)
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Given ε > 0, pick δ > 0 such that
x − x0 < δ
implies that
f (x, y) − f (x0 , y) < ε ∀y ∈ [a2 , b2 ]
where x0 ∈ I. Then, we see that
Z
Z
0
0
f (x0 , y) − f (x, y) < ε(b2 − a2 )
g(x ) − g(x) = f (x , y) − f (x, y) ≤
y∈[a,b]
y∈[a2 ,b2 ]
which is enough.
Now we claim the following.
Theorem 16.2 (Switching Order of Summation)
R
R
R
We have S f = x∈[a1 ,b1 ] g(x) = y∈[a2 ,b2 ] h(y).
Proof. Let
Z
F1 (t) =
g(x).
x∈[a1 ,t]
Let
Z
F2 (t) =
g(x).
x∈[a1 ,t]×[a2 ,b2 ]
If t = a1 , F1 (t) = F2 (t) = 0. So it suffices to show that F1 and F2 are differentiable and
have equal derivatives.
By the Fundamental Theorem of Calculus, F10 (t) exists and coincides with g(t).
For F2 it’s trickier. Recall that f is uniformly continuous, so for any ε > 0 we can find
h > 0 such that
∀y ∈ [a2 , b2 ] : f (t + h0 , y) − f (t, y) < ε
where 0 ≤ h0 < h.
We have
Z
Z
f (t, y) = h
(x,y)∈[t,t+h]×[a2 ,b2 ]
f (t, y)
t∈[a2 ,b2 ]
for obvious reasons (it’s just a slice of width h). To compute F20 (t), we put
Z
1
0
F2 (t) = lim
f (x, y)
h→0 h (x,y)∈[t,t+h]×[a2 ,b2 ]
#
"Z
Z
1
= lim
f (t, y) +
(f (x, y) − f (t, y))
h→0 h
(x,y)∈[t,t+h]×[a2 ,b2 ]
(x,y)∈[t,t+h]×[a2 ,b2 ]
Z
Z
1
=
f (t, y) lim
(f (x, y) − f (t, y))
h→0 h (x,y)∈[t,t+h]×[a2 ,b2 ]
y∈[a2 ,b2 ]
Z
1
(f (x, y) − f (t, y)) .
= g(t) + lim
h→0 h (x,y)∈[t,t+h]×[a2 ,b2 ]
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Evan Chen (Spring 2015)
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Hence we wish to check that the latter limit is zero. The following computation is
sufficient:
Z
1 lim
(f (x, y) − f (t, y))
h→0 |h| (x,y)∈[t,t+h]×[a2 ,b2 ]
Z
1
≤ lim
|(f (x, y) − f (t, y))|
h→0 |h| (x,y)∈[t,t+h]×[a2 ,b2 ]
1
< lim
|h| (b2 − a2 )ε
h→0 |h|
= ε(b2 − a2 ).
The result that we can switch summations is called Fubini’s Theorem.
§16.3 Open Sets
We’re done with boxes. Now let U be a bounded open
set in Rn , and let f : U → R be a
R
uniformly continuous function; we wish to define U f .
Recall that the support of f , supp(f ), is the closure of the nonvanishing points of f .
Assume first that supp(f ) is compact, and hence contained in some box S ⊆ Rn . Then
we define f˜ : Rn → R by setting f˜(x) to be f x on U and zero outside it. Then we define
Z
Z
f=
f˜.
U
S
more generally, if supp(f ) is not compact, we take a partition of unity {φk } as in last
lecture and let
Z
XZ
f=
f · φk .
U
k
U
We now need to verify that
• Re-ordering the terms of sum shouldn’t matter.
• If f has compact support, these two definitions should coincide.
• This doesn’t depend on the choice of {φk }
Well, the third bullet kind of implies the first and second. . . Although if partitions of
unity are indexed by a set rather than a sequence, then the first bullet point is a valid
concern and we get around it by following.
P R
Definition 16.3. We say f is integrable, that is, the series k f · φk is absolutely
convergent.
This follows, for example, that since f is bounded by M , since in that case
Z
X Z
XZ
X
f · φk ≤
|f | |φk | ≤
M · φk < ∞.
k
k
k
The second concern, again, follows by taking a trivial partition of unity with constant
value 1 on supp(U ). So let’s just do the third part.
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Evan Chen (Spring 2015)
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Lemma 16.4
If {φk }, {ψj } are two partitions of unity, then {φk ψj } is a partition of unity.
Proof. Exercise. OK fine I’ll write it out:
• Smoothness is clear.
• supp(φk · ψj ) = supp(φk ) ∩ supp(ψj ).
• Given neighborhoods Vx for the φ and Wx for the ψ, take the intersection.
• ToP
show the sum
realize there are only finitely many nonzero terms; the sum
Pis 1, P
is k,j φk ψj = k φk j ψj = 1.
Before we can get the final result, we need one more lemma.
Lemma 16.5 (Discrete Fubini)
P
Let k,j bk,j be absolutely convergent biseries. Then
X
bk,j =
XX
k,j
bk,j =
j
k
XX
j
bk,j .
k
This resolves the main result.
Theorem 16.6
If {φi }, {ψj } are partitions of unity, then
XZ
XZ
f φk =
f ψj .
j
k
Proof. We show both are equal to
XZ
f φk ψj =
k,j
P
X
j,k
R
f φk ψj . Observe that


XZ
X Z


f φk ψj =
f φk .
j
k
k
Similarly flipping the order of summation gives
XZ
XXZ
XZ
f φk ψj =
f φk ψj =
f ψj .
k,j
j
j
k
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Evan Chen (Spring 2015)
17 April 7, 2015
§17 April 7, 2015
R
Last time we saw how to define
Theorem that
Z
S
f for a box S. Also, for S = S1 × S2 , we saw Fubini’s
Z Z
Z Z
f=
f=
f.
S1 ×S2
S1
S2
S2
S1
Then, using the partitions of unity, we defined
Z
f
U
for U ⊆ Rn bounded, by summing over partitions of unity.
§17.1 Integration Over Vector Spaces
We want to define a notion
Z
f
U
for U ⊆ V . But this is not well-defined. For example, if V = R1 and f ≡ 1 then there’s
a change of basis called “multiply by V ” and we have
Z
Z
1 = b − a and
1 = c(b − a).
(a,b)
(ca,cb)
Lemma 17.1
T
f
Let Rn −
→ Rn −
→ R. Then
Z
U
f = |det T |
Z
T −1 (U )
f ◦ T.
Sketch of Proof. Suffices to show it for boxes, then note that boxes go to parallelpipeds.
What if it’s not linear?
§17.2 Change of Variables
The main theorem of today is the following.
Theorem 17.2 (Change of Variables Formula)
Let U1 and U2 be domains in Rn and suppose we have a setup
φ
f
U1 −
→ U2 −
→R
where φ is a diffeomorphism. Then
Z
Z
f=
U2
U1
|det Dφ | (f ◦ φ).
76
Evan Chen (Spring 2015)
17 April 7, 2015
Perhaps I should write this as
Z
Z
f (x) =
U1
U2
|det Dφ (x)| (f ◦ φ(x))
to emphasize the determine varies per point.
A quick remark. By using partitions of unity, we may assume that supp f is compact.
Also, by picking a proper cover, we can make nice assumptions on our set (since a proper
cover can be made subordinate to any cover).
Proof when n = 1. Using partitions of unity, we may assume U2 = (a2 , b2 ). Define
Z
F2 : (a2 , b2 ) → R by F2 (x) =
f
[∗,x]
where ∗ = a2 (or anything else), and let F1 = F2 ◦ φ. Thus F10 = φ0 · (F20 ◦ φ).
Since φ is a diffeomorphism, φ0 never vanishes. So WLOG we may assume φ0 > 0
(orientation-preserving); in that case φ(a1 ) = a2 and φ(b1 ) = b2 . (In the other case, a2
and b2 swap introducing a sign.)
By the Fundamental Theorem of Calculus,
Z
Z
f = F2 (b2 ) − F2 (a2 ) = F1 (b1 ) − F1 (a1 ) =
F10 .
(a2 ,b2 )
(a1 ,b1 )
Since F10 = φ0 · (F20 ◦ φ), meaning
Z
Z
f=
(a2 ,b2 )
(a1 ,b1 )
φ0 · (f ◦ φ)
as required.
Proof for n > 1 By Induction. Let 0 < m < n, so n = m + (n − m). We want to throw
Fubini at this.
Suppose we’re lucky enough that the following diagram commutes:
p1
-
where each pi is a projection
Rm
p2
φ U2
U1
pi : Ui ,→ Rn Rm
Define Um ⊆ Rm a domain. By Fubini’s Theorem, we may define F1 , F2 : Um → R by
Z
F2 (x) =
f (x, y)
y∈p−1
2 (x)∩U2
and
Z
F1 (x) =
y∈p−1
1 (x)∩U1
(f ◦ φ)(x, y) · |det Dφ(x, y)| .
Picture this as integrating over “slices”.
77
Evan Chen (Spring 2015)
17 April 7, 2015
U1
x
Rm = R1
Claim 17.3. F1 = F2 .
Proof. For all x, the map φ defines a diffeomorphism
φx = p1−1 (x) ∩ U1 → p−1
2 (x) ∩ U2 .
By induction, we know that
Z
F2 =
p−1
1 (x)∩U1
(f ◦ φ)(x, y) |det Dφx (x, y)| .
So, we just have to show
|det Dφ| = |det Dφx | .
This follows from the fact that we have the following commutative diagram
0
R
?
0
n−m
Dφx
∩
∩
?
Dφ - ?
Rn
?
?
?
id - m
R
Rn
R
R
?
n−m
m
?
?
?
0
0
Hence det Dφ = det id · det Dφx = det Dφx as needed.
So we’re done in that very fortunate case. More generally, since it suffices to doing
things locally, we claim we can reduce to this case as follows.
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Evan Chen (Spring 2015)
17 April 7, 2015
Claim 17.4. Let φ1 , φ2 , . . . , φn be the n projections of φ. For all x ∈ U1 there exists
an index i such that the map
φ̃i (x) = (φ1 (x), . . . , φi−1 (x), πi (x), φi+1 (x), . . . , φn (x))
is a diffeomorphism on a neighborhood of x.
Proof. We wish to use the Inverse Function Theorem. It suffices that
Dφ̃i (x) : Rn → Rn
is invertible. So we want to find an i so that the map
(Dφ1 (x), . . . , ei , . . . , Dφn (x))
is linearly independent. Such an i must exist since the Dφj are all linearly independent,
meaning it’s impossible that all the basis elements are linearly dependent with it.
This completes the proof.
§17.3 Integration Over Manifolds
Recall from PSet 8 that we have
5. For a real vector ` of dimension 1, we define an orientation on ` to be the choice
of one of the two cosets in ` \ {0} with respect to the equivalence relation
l1 ∼ l2 ⇐⇒ ∃c ∈ R>0 such that c · l1 = l2 .
For a finite-dimensional real vector space V , we define an orientation on V to be
an orientation on the 1-dim vector space Λn (V ), where n = dim(V ).
Let U ⊆ V be a domain. Let ω be a top-dimensional (meaning dim V -form, as dim V + 1
vanishes) bounded differential form on U , and set ε as an orientation of V . First, pick a
T
basis of V using Rn −
→ V . Then we define
Z
Z
ω=±
u,ε
T ∗ω
T −1 (U )
where the sign is +1 if T preserves the orientation and −1 otherwise. Here I mean to
look at
Λn T : Λn Rn → Λn V
and decide whether it preserves the orientation ε.
Lemma 17.5
This definition doesn’t depend on the choice of basis.
Proof. Equivalent to the lemma earlier about linear transformations.
I don’t want to bore you all to sleep, so I won’t prove it. I’ll put it on your
PSet. – Gaitsgory
Now let U ⊆ V .
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Evan Chen (Spring 2015)
17 April 7, 2015
Definition 17.6. An orientation of U is a continuous map
U → {NorthV , SouthV }
as topological spaces, where the range is a discrete space and its elements are the two
orientations of V . (Hence we assign an orientation to each connected component of U .)
Corollary 17.7
φ
Suppose U1 −
→ U2 is a diffeomophism of domains U1 ⊆ V1 and U2 ⊆ V2 . Let ω be a
top-dimensional form on U2 with compact support. Let εi be an orientation on Ui
for i = 1, 2. If ε1 and ε2 agree, then
Z
Z
ω=
φ∗ (ω2 ).
U2 ,ε2
U1 ,ε1
Here we say ε1 and ε2 agree under φ if for all x ∈ U1 , the map
∼
Λn Dφ(x) : Λn V1 −
→ Λn V2
agrees with ε1 and ε2 .
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Evan Chen (Spring 2015)
18 April 9, 2015
§18 April 9, 2015
§18.1 Digression: Orientations on Complex Vector Spaces
Let’s try to put orientations on complex vector spaces, because why not.
First, we put an orientation on the one-dimensional space C by regarding it as a
real vector space with basis (e1 , i · e1 ). We claim that for any choice of e1 , we can put
(e1 , i · e1 ) as a positive guy. To show that this is consistent, we need to show that the
linear transformation (e1 , i · e1 ) 7→ (e01 , i · e01 ) has positive determinant. One can check
this.
§18.2 Setup
Take our setup
φ
f
U1 −
→ U2 −
→R
and U2 domains in Rn and φ a diffeomorphism. We will assume that
(f ◦ φ) · |det Dφ|
which is true, for example, if we have compact supports. Then the change of variables
theorem stated that
Z
Z
(f ◦ φ) |det Dφ| =
f.
U1
U2
Recall that an orientation is a continuous map sending each x ∈ U to be εx an
orientation on U . Note that as an edge case, if U is a single point, then dim V = 0. We
define
Λ0 (V ) = R.
The reason for doing this is so that
Λtop (V1 ⊕ V2 ) ' Λtop (V1 ) ⊕ Λtop (V2 )
holds. But R has a preferred orientation already, so there is nothing to do in this edge
case.
Now recall how we defined
Z
Z
ω=
g
T −1 (U )
U,ε
for a domain U , a linear map T : Rn → V with pullback T ∗ (ω) = f dx1 ∧ . . . dxn , and
g = ±f depending on whether T preserves orientation. On the homework we show that
(a) this is independent of T , and
(b) if φ : U1 → U2 is a diffeomorphism and φ maps ε1 to ε2 , then
Z
Z
∗
φ (ω) =
ω
U1 ,ε1
U2 ,ε2
(with no constant factor!).
Let me try to explain the pullback a little more. Suppose φ : U1 → U2 is a diffeomorphism, and observe that we can put
φ = (φ1 , . . . , φn ).
81
Evan Chen (Spring 2015)
18 April 9, 2015
Suppose
ω = f dy1 ∧ · · · ∧ dyn
(different f here), where we’re using y for the coordinates in U2 . Then we calculate
φ∗ (ω) = φ∗ f φ∗ (dy1 ) · · · ∧ φ∗ (dyn )
= (f ◦ φ)d(y1 ◦ φ) ∧ · · · ∧ d(yn ◦ φ)
= (f ◦ φ)dφ1 ∧ · · · ∧ dφn
by definition; observe that we may expand
X ∂φj
dφ1 =
dyi .
yj
j
At the end of the day, we arrive at

∂φ1
∂y1
∂φn
∂y1
...
..
.
...

φ∗ (ω) = (f ◦ φ) det  . . .
∂φ1
∂yn
..


.  = (f ◦ φ) · det Dφ.
∂φn
∂yn
In the special case where T is a linear map, we have the nice property that det T can be
interpreted as the constant in Λn T : R → R.
i
Observe Dφ : Rn → Rn when viewed as a matrix in the standard basis has ∂φ
∂xi . Finally,
I remark that in case you haven’t noticed yet,
∂g def
= Dei g.
∂xi
You can think of this as saying that these pullbacks absorb the determinant; that’s
what they are for.
§18.3 Integration on Manifolds
Let X denote a differentiable manifold now, with an atlas
φα
Vα ⊇ Uα −→ Uα0 ⊆ X
of charts.
Definition 18.1. An orientation on X is the choice of an orientation εα on Uα for
every α with the following property: for all α or β, the diffeomorphism
φ−1
β ◦φα
0
0
0
0
Uα ⊇ φ−1
−−−−→ φ−1
α (Uα ∩ Uβ ) −
β (Uα ∩ Uβ ) ⊆ Uβ
preserves orientation.
As usual, we can run this definition on an atlas (a set of charts which cover every
point), and it will extend uniquely to an orientation on the maximal atlas of X.
A second definition is as follows.
An orientation amounts to a choice of an orientation εx on Tx X, for each x ∈ X. This
assignment needs to be “continuous”. To talk about continuity, we resort to charts: we
want it to be continuous on every chart via
φ−1
α
(Dφα )−1
Uα0 −−→ Uα ,→ Vα −−−−−→ Tx X.
Finally we can define the integral (ack).
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Evan Chen (Spring 2015)
18 April 9, 2015
Definition 18.2. Let X be a differentiable manifold, and ω a compactly supported
top-dimensional differential form on X, and ε an orientation on X. Select a partition
of unity Ψα for each chart (Uα , φα ). (Since the support of ω is compact, there are only
finitely many α’s such that the support of φα intersects the support of ω). Then we
define
Z
XZ
def
ω =
φ∗α (ω · ψα ).
X,ε
Uα ,εα
α
Proposition 18.3
This is independent of the choice of atlas.
φ
φ
Proof. Let (Uα −
→α Uα0 ) and (Uβ −
→β Uβ0 ) be two totally unrelated atlases equipped with
partitions of unity ψα and ψβ . We compute
XZ
X
XZ
∗
φα (ψα · ω) =
φ∗α (ψα · ω ·
)
α
Uα ,εα
α
=
Uα ,εα
XXZ
α
β
Uα ,εα
β
φ∗α (ψα · ω)
where the summation is okay due to there being finitely many zero terms. So now we
wish to check that
Z
Z
∗
φα (ψα · ω) =
φ∗β (ψβ · ω)
Uα ,εα
Uβ ,εβ
in which case we’ll be done by symmetry.
φβ
φα
Uα −→ Uα0 ⊆ X ⊇ Uβ0 ←− Uβ .
Notice that the support of ωψα ψβ is contained in Uα0 ∩ Uβ0 , since ψα has support inside Uα0
−1
and ψβ has support inside Uβ0 . Effectively we’re integrating over φ−1
α (Uβ ) and φβ (Uα ).
This is done using the transition function
φ−1
β ◦φα
φ−1
−−−−→ φ−1
α (Uβ ) −
β (Uα ).
§18.4 Stoke’s Theorem
Let S ⊆ Rn be a box contained inside some domain U . Let ω ∈ Ωn−1 (U ) be the set of
continuously differentiable n − 1 forms.
Now we state Stoke’s Theorem, which is the generalization of the Fundamental Theorem
of Calculus.
We can consider dω, which is now a top-dimensional differentiable form. On the other
hand, we can consider ∂S, which is the “boundary” of the box, and then take its interior
˚ (In other words, take the union of the interiors of the faces.) This is a
to get ∂S.
manifold.
83
Evan Chen (Spring 2015)
18 April 9, 2015
Example 18.4
˚ consists of four open segments.
If S is a square, then ∂S
R
R
We would now like to assert that S̊ dω is equal to ∂S
˚ ω. But to do this we need
orientations. For dω this is no problem as we have a standard orientation on Rn . But it
˚
remains to orient ∂S.
Well, given a vector space V = V1 ⊕ V2 , an orientation of Vi can be specified from an
˚ as follows: let V1
orientation of V2 and V (the order of this matters). Thus, we orient ∂S
be the orthogonal one-dimensional vector out of a face, and V2 the rest of the face. We
take the orthogonal guy out of the face in V1 as the positive direction, plus the standard
orientation on Rn , to get the orientation on each face.
+
Theorem 18.5 (Baby Stoke’s Theorem)
With the orientations above,
Z
Z
dω =
S̊
˚
∂S
ω.
Proof. First, we handle the case n = 1. Thus, we have S = [a, b]. Then, ω ∈ Ω0 (R) is
represented by a single function f , and
Z
Z
ω=
f 0 dx = f (b) − f (a)
S̊
[a,b]
by the Fundamental Theorem ofR Calculus.
R
Hence, we just need to check S̊ ω equals this. Observe that S̊ = {a, b}. The point b
has orientation agreeing with that of R, and the point a has orientation against that of
R. So we get f (b) and f (a), yay.
“Do you want me to jump straight to the inductive case or do n = 2 first?” –
Matt
“Which do you want” – Gaitsgory
“n = 6.” – Matt
(laughter)
“Do n = 120!” – James
We’ll do n = 2; the case n ≥ 2 is homework.
We have ω ∈ Ω1 (R2 ), so we may put
ω = f1 dx1 + f2 dx2 .
The positive orientations are given as follows.
84
Evan Chen (Spring 2015)
18 April 9, 2015
(a, d)
(b, d)
(a, c)
(b, c)
First, consider ω = f dx1 for simplicity (the other guy is analogous and we’ll be done
by adding). We compute
dω =
∂f
∂f
dx2 ∧ dx1 = −
dx1 ∧ dx2 .
∂x2
∂x2
Now we compute this using Fubini’s Theorem:
Z
Z
∂f
dω = −
∂x2
S̊
ZS̊
Z
∂f
=−
s∈[a,b] t∈[a,b] ∂x2
Z
=−
(f (s, d) − f (s, c))
s∈[a,b]
Z
Z
=
f (s, c) −
f (s, c)
s∈[a,b]
s∈[a,b]
where the last step is by the Fundamental Theorem of Calculus. This gives us the two
vertical segments, and the horizontal segments contributes nothing since x1 is constant
on them, and dx1 = 0.
Adding in f dx2 gives the other contribution. Let’s make sure we didn’t botch the
signs. Put ω = f dx2 , and
∂f
dω =
dx1 ∧ dx2 .
∂x1
Z
Z
dω =
S̊
ZS̊
∂f
∂x1
∂f
t∈[c,d] s∈[a,b] ∂x1
Z
Z
=
f (b, t) −
Z
=
t∈[c,d]
t∈[c,d]
This matches our orientation sanity checks.
85
f (a, t).
Evan Chen (Spring 2015)
19 April 14, 2015
§19 April 14, 2015
Theorem 19.1
Let ω ∈ Ωn−1 (U ) such that B(x, r) ⊆ U , then
Z
Z
dω =
ω.
∂B(x,r)
B(x,r)
Let X ⊇ ∂X be Hausdorff, second countable, and paracompact.
Definition 19.2. A structure on (X, ∂X) of an n-dimensional manifold with boundary
ϕα
is a supply of charts R≥0 × Rn−1 ⊇ Uα ∼ Uα0 ⊆ X such that ϕα (∂X) = Uα ∩ ({0}) × Rn−1
and a compatibility condition.
We must first define what it means for a map of half spaces to be C ∞ . Take ϕ a
homeomorphism.
∂U1
⊆
U1
⊆
R≥0 × Rn−1
⊆
R≥0 × Rn−1
ϕ
∂U2
⊆
U2
◦
◦
We may decompose ∂U ⊆ U ⊇ U , with ∂U = U ∩ ({0} × Rn−1 ), U = U ∩ (R>0 × Rn−1 ).
Then we want ϕ to be C ∞ on the interior, and for every x ∈ U1 , there exists r > 0 and a
C ∞ extension of ϕ to B(x, r). From here, we have all of our old theorems from manifolds.
The only difference is that there is a boundary.
Example 19.3
Consider B(x, 1). We claim that we can construct a chart between the strictly upper
half sphere and (0, 1] × Rn−1 by projecting in the usual manner, with a proportion
of depth related to the distance of each point in the circle from the center. (Aaron
says the formula is dividing by the secant of the angle.) We can then grab the center
point just by considering the open ball.
Let (X, ∂X) be a manifold with boundary, f : X → R.
Definition 19.4. f is C ∞ if for some/any chart, f | ◦ is C ∞ and for every x there is r
X
such that f extends to a C ∞ function on B(x, r).
Definition 19.5. A k-differential form on (X, ∂X) is (on some/any chart)
◦
• ω on X
• For every x there exists r such that ω extends to a C ∞ function on B(x, r).
We then want to define a restriction to the boundary, Ωk (X) → Ωk (∂X). Fix x ∈ ∂X,
and a chart that contains X. Extend φ∗ (ω) to B(x, r) and then we can just restrict to
B(x, r) ∩ Rn−1 . We could have different extensions of φ∗ (ω), however, the restriction will
be the same by continuity. That is, for ω1 , ω2 : B(x, r) → Λk (Rn ), ω1 |B(x,r)×(R>0 ×Rn−1 ) =
ω2 |B(x,r)×(R>0 ×Rn−1 ) , so by continuity, they are also equal on the boundary.
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Evan Chen (Spring 2015)
19 April 14, 2015
◦
Definition 19.6. An orientation on (X, ∂X) is an orientation on the interior X.
Given (X, ∂X), compactly supported ω, and an orientation , we want to define
Using partitions of unity we can assume supp(ω) ⊆ Uα . Then
Z
Z
ω := ◦ φ∗α (ω).
Uα ,
R
X, ω.
U α ,
◦
Given an orientation on X, we then must define an orientation on the boundary. Let
x ∈ ∂U , we have the short exact sequence
0 → Rn−1 → Rn → Rn /Rn−1
splits so
Λn (Rn ) ' Rn /Rn−1 ⊗ Λn−1 Rn−1 .
Thus is suffices to orient Λn (Rn ) and Rn /Rn−1 . We just need to orient Rn /Rn−1 since
we already have the former, and here we just take out to be positive.
Digression on “out”: Let γ : (−, ) → U with γ(0) = x a path such that Dγ ∈ Rn ,
that projects to a positive vector in Rn /Rn−1 .Then for t > 0 sufficiently small, γ(t) 6∈ U .
Theorem 19.7 (Stoke’s Theorem)
Let (X, ∂X) be a manifold with boundary, ω ∈ Ωn−1 (X) with compact support, and
orientation on X. Then
Z
Z
dω =
ω.
∂X,ε
X,ε
P
Proof. Take a partition
of
unity,
ω
=
⊆ Uα .
α ωψα . We can therefore assume supp(ω)
R
R
R
∗
∗
We want to show ◦ φ (dω) = ∂Uα , φ ω. Note that the LHS is equal to ◦ dφ∗ (ω).
U α ,
U α ,
Then we can just draw a box around our function and declare it to be 0 on the outside,
and we’re done by last class.
R
Note that supp(ω) ∩ ∂X = ∅ =⇒ X, dω = 0 immediately by Stoke’s Theorem.
Dealing with the case of manifolds with corners will be on the homework.
[Gaitsgory is saying that the case of manifolds with corners is not class level
difficulty, it’s only homework level]
Stefan: “I feel like the ‘homework level’ stuff is getting harder and harder
over time.”
Gaitsgory: “Well... You’re getting older.”
87
Evan Chen (Spring 2015)
20 April 16, 2015
§20 April 16, 2015
Finally complex analysis today. Most of these notes are excerpted from my Napkin
project, see http://web.evanchen.cc/napkin.html.
§20.1 Complex Differentiation
Let f : U → C be a complex function. Then for some z ∈ U , we define the derivative
at z0 to be
f (z0 + h) − f (z0 )
lim
.
h→0
h
Note that this limit may not exist; when it does we say f is differentiable at z0 .
This is not the same as being differentiable R2 → R2 , since the latter is
f (x + v) − f (x) − T (v)
→ 0.
kvk
Complex multiplication is a special case of T , rather than vice-versa. In fact, T “comes
from λ” if and only if T ◦ (i · −) = (i · −) ◦ T .
Let f : U → C be differentiable, and write f (x + iy) = fR (x + iy) + i · fI (x + iy). Then
we see the total derivative of f ought to be
!
∂fR
∂x
∂fI
∂x
∂fR
∂y
∂fI
∂y
.
The fact that it commutes with i ◦ − means that
!
∂fR
∂fR
∂fR
0 −1
∂x
∂y
∂x
=
∂fI
∂fI
∂fI
1
0
∂x
∂y
∂x
∂fR
∂y
∂fI
∂y
!
0 −1
1 0
Now we multiply the matrices.
If you fail to do it, I’ll really know I taught you well.
This gives:
Corollary 20.1 (Cauchy-Riemann Equations)
If f is complex differentiable it satisfies
∂fI
∂fR
=−
∂y
∂x
and
∂fR
∂fI
=
.
∂x
∂y
Moreover if the partials are continuously differentiable then the converse is true as
well.
These are really strong.
But note that having a complex differentiability is actually much stronger than a real
function having a derivative. In the real line, h can only approach zero from below and
above. and for the limit to exist we need the “left limit” to equal the “right limit”. But
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Evan Chen (Spring 2015)
20 April 16, 2015
the complex numbers form a plane: h can approach zero from many directions, and we
need all the limits to be equal.
So we restrict our attention to differentiable functions called holomorphic functions. It
turns out that the multiplication on C makes all the difference. The primary theme in
what follows is that holomorphic functions are really, really nice, and that knowing tiny
amounts of data about the function can deterimne all its values. Some highlights:
• It’ll turn out that having a first derivative is enough to guarantee having all
derivatives (which is unlike the real case, in which no number of derivatives will
guarantee the rest).
• Now that you have all derivatives, you can take the Taylor series, (unlike the real
case, in which even if you have all derivatives it’s possible the Taylor series is totally
wrong).
• It’ll turn out that knowing just the values of a holomorphic function on the boundary
of the unit circle will tell you all the values everywhere.
• Even knowing just the values of the function at 1, 12 , 13 , . . . is enough to determine
the whole function!
• Integrals in the complex plane will just work like magic.
If a function f : U → C is complex differentiable at all the points in its domain it is
called holomorphic. In the case U = C, we sometimes call the function entire.
§20.2 Examples
Example 20.2 (Not Differentiable Function)
Complex conjugation, z 7→ z, is an example of a map which is not differentiable at
zero. As we approach 0 from various directions, we get different values of the ratio
f (h)
h vary.
For example, f (z) = 6, f (z) = z 6 work.
“It’s not just ‘like 1’, it is 1” – Gaitsgory
Example 20.3 (Examples of Holomorphic Functions)
In all the examples below, the derivative of the function is the same as in their real
analogues (e.g. the derivative of ez is ez ).
(a) Any polynomial z 7→ z n + cn−1 z n−1 + · · · + c0 is holomorphic.
(b) The complex exponential exp : x + yi 7→ ex (cos y + i sin y) can be shown to be
holomorphic.
(c) sin and cos are holomorphic when extended to the complex plane by cos z =
iz
−iz
eiz +e−iz
and sin z = e −e
.
2
2
(d) As usual, the sum, product, chain rules and so on apply, and hence sums,
products, nonzero quotients, and compositions of holomorphic functions are also holomorphic.
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You are welcome to try and prove these results, but I won’t bother to do so.
Example 20.4
f (z) = 1/z is complex differentiable for f : U → C → 0.
“Oh, it still says theorem. It’s not ‘theorem’, that’s a joke. It’s ‘examples’.” –
Gaitsgory
Note ez =
zk
k k!
P
works here.
“Do you remember when we did applied math between those two big snows?”
– Gaitsgory
§20.3 Inverse Function Theorem
Still applies since the inverse of a nonzero λ·− : C → C is still just complex multiplication.
§20.4 Complex Differentiable Forms
For U ⊆ Rn , we can consider
Ωn (U ) ⊗R C
and view it as a formal pair ωR + iωI . Thus
Ω•C (U )
is a ring, and we can speak of the differential d : ΩnC (U ) → Ωn+1
C (U ) as before. Also we
can take a pullback as before.
From now on, we will assume that all our holomorphic functions are continuously
differentiable. This is not needed, but it will make life easier.
§20.5 Cauchy Integral Formula
Let z0 ∈ U ; then we can consider disks in the plane.
Take z ∈ Ω0C (U ), say z = x + iy. Then
dz = dx + idy ∈ Ω1C (U ).
“Now I’m going to multiply them.”
“You can write formal symbols without understanding what they mean.”
– Gaitsgory
So the point of dz = dx + idy means that to integrate, we just sum the real and
imaginary parts separately.
Now, suppose we want to integrate
Z
f (z)
dz
z
S(z0 ,r) − z0
where we pick the counterclockwise orientation; thus it should give us some complex
number.
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Theorem 20.5 (Cauchy’s Integral Formula)
For a holomorphic function f , let S be a circle of any radius such that z0 is inside S.
Then
Z
f (z)
dz = 2πif (z0 ).
S z − z0
Everything about holomorphic functions is going to follow from this.
Lemma 20.6
Let f be a complex continuously differentiable function. Then f is holomorphic if
and only if
d(f dz) = 0.
Proof. Let f = fR + ifI . Then quick computation gives
∂fR
∂fI
∂fI
∂fR
d ((fR + ifI )(dx + idy)) =
dy ∧ dx −
dx ∧ dy +
dy ∧ dx −
dx ∧ dy i.
∂y
∂x
∂y
∂x
This is zero if and only if the Cauchy-Riemann equations.
“How many terms are there altogether? Eight. That’s a large number.” –
Gaitsgory
Corollary 20.7
The value of the integral in the problem in Cauchy’s Integral Formula doesn’t depend
on the circle chosen.
Proof. Consider two circles as described. Let
ω=
f (z)
dz.
z − z0
We invoke Stoke’s Theorem, where the manifold in question is the annulus X cut out by
these two circles. This gives
Z
Z
dω =
ω.
X,ε
∂X
But ω is holomorphic, hence dω = 0. Also, ∂X is the two circles, so
Z
Z
Z
0=
ω=
ω−
ω.
∂X
outer
inner
Of course, these don’t actually have to be circles.
§20.6 Computation when f = 1
We still want to prove Cauchy’s Integral Formula. There’s a π, so we need to do something.
“Rachit, what’s π?” –Gaitsgory
Now we prove Cauchy’s Integral Formula. First, we prove it in the case z0 = 0,
f (z) = 1, and S the circle of radius 1.
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“In the past weeks we talked about integrals and partitions of unity and
whatnot. . . So now we have to actually do it.” – Gaitsgory
We have the function
R → S ⊆ R2 by t 7→ exp(it).
Let φ(t) = exp(it).
Claim 20.8. φ∗ ( dz
z ) = i dt.
Proof. We have
φ∗ dz
d(z ◦ φ)
(z ◦ φ)0 dt
=
=
.
z◦φ
z◦φ
z◦φ
Since z ◦ φ = exp(it), its derivative is just i exp(it), so we’re done.
Claim 20.9.
Z
S 1 ,ε
dz
=
z
Z
φ
∗
[0,2π]
dz
z
.
Proof. Clear, modulo an issue of [0, 2π) instead of [0, 2π]
Then
Z
φ
∗
[0,2π]
dz
z
Z
=
idt = 2πi.
[0,2π]
§20.7 General Proof of Cauchy’s Formula
Now we want to show the general case
Z
Z
f (z)
f (0)
= 2πi · f (0) =
.
z
Sr
Sr z
This amounts to
Z
Sr
f (z) − f (0)
dz = 0.
z
Taking the pullback again, the left-hand side equals
Z
Z
∗ dz
φ
((f (z) − f (0)) ◦ φ) = i ·
f (r exp(it)) − f (0).
z
[0,2π]
[0,2π]
But r is arbitrary, and we can make f (r exp(it)) − f (0) as small as we want by continuity.
More explicitly, for any ε > 0 we can take r small enough so that
|f (r exp(it)) − f (0)| < ε
whence
Z
ε→0
f (r exp(it)) − f (0) < 2πε −−−→ 0.
[0,2π]
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§21 April 21, 2015
Let z0 be a point in a domain U . Then we showed last class that
Z
1
f (z)dz
f (z0 ) =
,
2πi S z − z0
where f is complex differentiable and C 1 .
Theorem 21.1 (Maximum Modulus Principle)
Let S be a circle centered at z0 . Then |f (z0 )| ≤ maxz∈S |f (z)|.
Proof. For simplicity, set z0 = 0, and
Integral Formula. Then
we can
R
R f (z)
R use the Cauchy
take [0, 2π] → S by t 7→ exp(it), so S f (z)
=
i
f
(exp(it)).
Then
|
dz
[0,2π]
S z dz| =
R
R
Rz
| [0,2π] f (exp(it))| ≤ [0,2π] |f (exp(it))| ≤ [0,2π] max |f (z)| = 2π maxz∈S |f (z)|.
Theorem 21.2 (Liouville)
Let f be holomorphic on all of C (that is, f is entire), and for every there is R
such that |f (z)| < for |z| > R. Then f = 0.
Proof. Fix z0 , by Theorem 21.1 we can bound f (z0 ) < for every , thus f = 0.
Theorem 21.3 (Fundamental Theorem of Algebra)
Every complex polynomial has a root.
1
Proof. Suppose there is not a z such that p(z) = 0. Then consider p(z)
is well defined on
C. Moreover, at large z, the highest term of the polynomial dominates the function, so
1
1
the p(z)
satisfies the conditions of Liouville’s Theorem. Thus p(z)
= 0, contradiction.
Theorem 21.4 (Holomorphic Functions are Smooth)
Let f be holomorphic and C 1 on U . Then f is C ∞ .
Proof. It suffices to show fR0 is C 1 by problem 8 on the PSet. Take any ball B ⊆ U
(z)dz
R
with z0 ∈ B, then f (z0 ) = S fz−z
. Partials exist by the midterm, as we have ∂f
∂x0 =
0
R
1
∂
1
2πi
∂x0 ( z−z0 )f (z)dz, and the inner term is clearly differentiable with respect to z0 .
Then it continues to satisfy the Caucy-Riemann equations after you do the integration,
which finishes the problem.
“We’ll now state the following false theorem.” –Gaitsgory
Let f be C ∞ on (a, b) such that for some x ∈ (a, b), f (n) = 0. Then f = 0.
1
We’ve already seen this is false–consider for instance e− x at x = 0.
“This is completely false in the real world.” – Gaitsgory
However, it is true for holomorphic functions.
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Theorem 21.5 (Holomorphic Functions are Analytic)
Let f be holomorphic on U and B ⊆ U a disk of radius r, with z0 the center. Then
f (z) =
X f (n) (z0 )
n!
(z − z0 )n
uniformly and absolutely on B.
Proof. By the Cauchy Integral Formula, f (z) =
larger than B.
1
2πi
R
f (e
z)
S (e
z −z) ,
where S is a circle slightly
“No, z 0 isn’t good, what should we use... Ah, ze.” –Gaitsgory (paraphrased)
Then
1
1
z z2
= (1 + + 2 + ...)
ze − z
ze
ze ze
converges uniformly and absolutely, by taking the radius of S to be r0 , so that
r
z
< 1.
| |<
ze
r0
R
R
Then gn → g uniformly implies S gn → S g, so we are left with
Z
∞
f (e
z )de
z X zn
(
),
ze
zen
n=m
which
want to bound < , which is clear. This will show that the sequence
Pn Rwef (e
z) i
(
i=0 S zei+1 )z converges to f (z).
R
(n)
1
1
Now f n!(0) = 2πi
f (e
z )de
z · ( ze−z
)(n) |z0 =0 , which follows by the Cauchy formula on
0
f (n) (0).
Theorem 21.6
p
Let ai be P
a sequence of complex numbers such that lim sup n |an | ≤ r10 . Then
the series
an z n converges pointwise to a holomorphic function on Br , and the
convergence is uniform on B r , for any r < r0 .
Proof.
First, uniform convergence on B r : Take r < r0 < r0 , then for all but finitely many
p
n, n |an | < r10 =⇒ |an | < (r10 )n , so |an z n | < ( rr0 )n , which we know how to estimate.
P
r n
r m
1
Then the tails are ≤ ∞
n=m ( r0 ) = ( r0 ) ( 1− r0 ).
r
We now need a lemma. In particular, if fn → f uniformly on some domain in U , with
fi holomorphic, then we’d like to claim f is as well. This will be presented below. Then
we can apply this to our case to get that our function is holomorphic, and then by writing
Br0 = ∪r<r0 Br , we get our theorem. (Note that our functions are holomorphic on B r , so
a fortiori they are holomorphic on Br .)
We used the following lemma in our theorem:
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Lemma 21.7
Let fn → f uniformly on U be holomorphic. Then f is holomorphic.
1
Proof. We have f (z0 ) = lim fn (z0 ) = lim 2πi
limit is a holomorphic function.
R
dz
S z−z0 fn (z)
=
1
2πi
R
dz
z−z0 f (z),
hence the
Theorem 21.8 (Identity Theorem)
Let U0 ⊆ U and f0 a holomorphic function on U0 . Assume U is connected. Then
there is at most one f on U such that f |U0 = f0 . (We say that there is at most one
analytic continuation of f0 .)
Proof. Let f1 , f2 be two holomorphic extensions, then (f1 − f2 )|U0 = 0. We may therefore
assume that f0 = 0, and we wish to show f = 0. Set V n = {z ∈ U | f (n) (z) = 0}. Note
these are closed sets. Then write V = ∩n V n , and the intersection of closed sets is closed,
so V is closed as well.
We claim that V is also open in U . Given z ∈ V , there is a disk B ⊆ U around z.
P f (n) (z0 )
On B, f (z) =
(z − z0 )n = f = 0 on all of B, so V contains B =⇒ V is open.
n!
Then U connected implies V = U .
On Thursday we will study the geometry of holomorphic mappings. Suppose U1 , U2
f
are domains, and U1 −
→ U2 . Then f is biholomorphic if f, f −1 are both holomorphic.
Example 21.9
Take U1 = C and U2 = B. There are diffeomorphisms between C and B through
stereographic projections. However, we’ll prove that there is not a biholomorphic
map between the two.
Theorem 21.10 (Liouville)
A bounded holomorphic function on C is constant.
From this, we can conclude C 6= B.
Example 21.11
Take U1 = C and
U2 = {z ∈ C | Im z > 0} .
Here U2 is holomorphic to B by the map exp(iz), so U2 = B 6= C holomorphically.
We will also show that all transforms of C are of the form az + b. Similarly, U2 from
Example 21.11 (the upper half plane) may be manipulated exclusively by maps of the
form z 7→ az+b
.
bz+a
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§22 April 28, 2015
Additional office hours tomorrow, Wednesday, 4PM-5PM.
The final will have
1. Differential equations (vector fields, commutator)
A. A bonus from the preceding pset
2. Orientations
3. Complex Analysis
4. Complex Analysis
B. Complex Analysis
§22.1 Riemann Mapping Theorem
Today we prove the Riemann mapping theorem.
Theorem 22.1 (Riemann Mapping Theorem)
Let U be a simply connected domain of C which is not C. Then there exists a
biolomorphic function
f : U → B(0, 1).
Proof. First, consider the set S of injective functions f : U → B(0, 1). Pick any z0 ∈ U .
Consider the map
S → R≥0 bys 7→ f 0 (z0 ) .
Let F be the element of S such that
0
F (z0 ) ≥ |f (z0 )| .
Note that we haven’t checked that F exists or even that S is nonempty, but taking that
on faith for now, we have:
Claim 22.2. F (z0 ) = 0 and F is surjective.
“Those of you who have not lost the ability to differentiate as a result of
55. . . ” – Gaitsgory
“Will we have to differentiate on the final exam?”
“OH YEAH”
Proof. For the first part, assume for contradiction F (z0 ) 6= 0. Then set
G(z) =
F (z) − F (z0 )
1 − F (z0 )F (z)
and verify that G ∈ S still. Then
G0 (z0 ) =
F 0 (z0 )
.
1 − |F (z0 )|2
But |G0 (z0 )| > |F 0 (z0 )|, contradicting the fact that F is maximal.
Now we show it’s surjective. Assume w isn’t in the image.
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“What letters of the alphabet do we know? We know F , we’ve used it, we’ve
used G. . . ”
F (z)−w
Let H0 (z) = 1−w·F
(z) . Since this function is non-vanishing, and the domain is simply
connected, we can find a function H1 such that H1 (z)2 = H0 (z) (“take square roots”).
Since H0 is injective so is H1 , thus H1 ∈ S too. Then let
H2 (z) =
H1 (z) − H1 (z0 )
1 − H1 (z0 ) · H1 (z)
.
With a very fun computation we get
0
+ |w|
H2 (z0 ) = 1 p
· F 0 (z0 )
2 |w|
By AM-GM, we have
which is the desired contradiction.
1 + |w|
p
>1
2 |w|
Missing
figure
James Tao’s diagram
Next, we will show the following lemma.
Lemma 22.3
Let f : U → C be holomorphic and injective. Then f 0 (z) 6= 0 for all z ∈ U .
Not true in R, take f (x) = x3 for example.
Proof. Assume not. By shifting, we have f (0) = f 0 (0) = 0. Write f (z) = z n g(z) for some
nonvanishing g. Restricted to a small disk, we can extract an nth root of g (according to
the last pset), so we may put f (z) = (z · h(z))n for some on some small neighborhood of
0. We claim z · h(z) is biholomorphic on a small disk; indeed its derivative is h(0) 6= 0 on
a small disk.
So on a small disk, we have
(−)n
z·h(z)
disk −−−→ disk −−−→ C
is injective, but the first arrow is an isomorphism and the second arrow is not injective,
which is impossible.
Now, non-vanishing differential is sufficient to get that f is biholomorphic. That’s all.
Oops no actually so we need to show the F exists in the first place. First, we show
that S 6= ∅. We have a function
U → B(0, 1).
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Since U 6= C, there exists a ∈
/ C. So we can take
√
z−a
which is nonvanishing and holomorphic on U . Let φ1 and φ2 be these two square roots.
Claim 22.4. Let i ∈ {1, 2}. If φi (z 0 ) = φi (z 00 ) then z 0 = z 00 .
Proof. Trivial, do it yourself.
Now φ1 and φ2 have nonvanishing derivative, so they have open images. Let w be in
the image of φ1 , arbitrarily. Then there exists r such that B(w, r) is contained inside the
image of φi , but does not intersect the image of φ2 . Set
f (z) =
r
.
w − φ2 (z)
This f works.
Finally, we wish to show that S indeed has a maximum as we claimed. For this we
appeal to the following.
Theorem 22.5
Let S be any set of holomorphic functions on U such that there exists a Λ such that
|f (z)| ≤ Λ such that for all f ∈ S and z ∈ U . Then any sequence fn ∈ S has a
subsequence that converges uniformly for all K ⊆ U .
By Morera’s Theorem, the limit of the uniform convergence is actually holomorphic.
Theorem 22.6 (Hurwitz)
Let fn → f uniformly on every compact, where f is nonconstant. If all the fi are
injective then so is the limit f .
Together these imply the result.
§22.2 Proofs of First Theorem
First, we prove the first theorem. Note that since an open set can be written as a union
of nestned compacts, it suffices to do this for a given compact K. Check ∃r such that
∀x ∈ K, B(x, r) ⊆ U .
Claim 22.7. fn are equicontinuous when restricted to K.
Proof. Blah.
§22.3 Proof of Hurwitz’s Theorem
It suffices to show the following:
Theorem 22.8 (Hurwitz’)
If fn → f uniformly on U 3 z0 and f (z0 ) = 0 but f 6≡ 0, then for all r there exists
N such that for all n ≥ N , fn will have a zero on B(z0 , r).
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“‘holomolomorphic”
Theorem 22.9
Let g be holomorphic on U and let B(z0 , r) ⊆ U . Assume g has no zeros on ∂B(z0 , r).
If
Z
g 0 (z)
dz 6= 0
∂B(z0 ,r) g(z)
then g has zeros in the ball.
Proof. If g has no zeros then Cauchy’s Theorem shows that the integral is zero.
Then we can just note that fn → f means fn0 → f 0 and then
Z 0
Z 0
fn
f
dz →
dz.
fn
f
This establishes Hurwitz’ and hence Hurwitz.
§22.4 A Theorem We Tried to Use But Couldn’t
We deduce the above from
Theorem 22.10
Let f and g be holomorphic on U such that |g| < |f |. Then if f has a zero in B(z0 , r)
then so does f + g.
Proof. Write f + g = f · (1 + fg ).
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